Ch. - 15 Probability

Ch. - 15 Probability

Chapter 15 Ex.15.1 Question 1

Complete the following statements:

(i) Probability of an event \(\rm{E +}\) Probability of the event 'not \(\rm{E}\)'__________.

(ii) The probability of an event that cannot happen is __________ Such an event is called __________.

(iii) The probability of an event that is certain to happen is __________Such an event is called __________.

(iv) The sum of the probabilities of all the elementary events of an experiment is  __________.

(v) The probability of an event is greater than or equal to \(0\) and less than or equal to __________.

 

Solution

Video Solution
 

 

Steps:

(i) Probability of an event \(\rm{E +}\) Probability of the event 'not \(\rm{E}\)'\(\begin{align}\underline{=1}\end{align}\)

(ii) The probability of an event that cannot happen is \(\underline{0.}\) Such an event is called impossible event.

(iii) The probability of an event that is certain to happen is \(\underline{1.}\) Such an event is called sure event.

(iv) The sum of the probabilities of all the elementary events of an experiment is  \(\underline{1.}\)

(v) The probability of an event is greater than or equal to \(\underline{0}\) and less than or equal to \(\underline{1.}\)

Chapter 15 Ex.15.1 Question 2

Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

 

Solution

Video Solution

 

(i) A driver attempts to start a car. The car starts or does not start?

Steps:

It does not have equally likely outcomes because the probability of it depends on many other technical factors.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot?

Steps:

It does not have equally likely outcomes because the probability here depends on the ability of the player.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

Steps:

It has equally likely outcomes because the probability is equal in both the cases, it can be true or false.

(iv) A baby is born. It is a boy or a girl?

Steps:

It has equally likely outcomes because the probability is equal in both the cases, it can be a boy or girl.

Chapter 15 Ex.15.1 Question 3

Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

 

Solution

Video Solution
   

 

Steps:

Tossing a coin is considered to be a fair way of deciding which team should get the ball at the beginning of a football game because it has equally likely outcomes, it has only two outcomes either head or tail, so the probability for both the team is equal.

Chapter 15 Ex.15.1 Question 4

Which of the following cannot be the probability of an event?

(i)  \(\begin{align} \frac23 \end{align}\)       

(ii) \(\begin{align} -1.5 \end{align}\)      

(iii)  \(\begin{align}15 \% \end{align}\)       

(iv) \(\begin{align}0.7 \end{align}\)

 

Solution

Video Solution

 

Steps:

We know that the probability of an event lies between \(0\) and \(1\) i.e. \(0 \le{ P(E)} \le 1\) and it cannot be less than \(0\) and greater than \(1.\)

So, (ii) i.e. \(-1.5\) cannot be the probability of an event because it cannot be negative.

Chapter 15 Ex.15.1 Question 5

If \(\begin{align}{P(E)} = 0.05,\end{align}\) what is the probability of 'not \({}E\)?

 

Solution

Video Solution

 

What is known?

\(P(E) = 0.05\)

What is unknown?

Probability of ‘not E’

Reasoning:

Probability of not happening an event is equal to \(1\)  minus probability of happening an event.

Steps:

\[\begin{align} {P}({E}) &=0.05\end{align}\]

We know that

\[\begin{align} {P}\,({E})+{P}\,(\text { not } {E}) &=1 \\ 0.05+{P} (\text { not }E) &=1 \\ {P}\,(\text { not } {E}) &=1-0.05 \\ {P}\,(\text { not }E) &=0.95 \end{align}\]

Thus the probability of  'not \({E}\)' is \(0.95.\)

Chapter 15 Ex.15.1 Question 6

A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out

(i) an orange flavored candy?

(ii) a lemon-flavored candy?

 

Solution

Video Solution

 

What is known?

Bag contain lemon flavoured candies.

What is unknown?

(i) an orange flavoured candy.

(ii) a lemon flavoured candy.

Reasoning:

We can check whether the event is Possible or Impossible

Steps:

(i) A bag contains only lemon flavored candy so there is no probability to take out orange flavored candy.Therefore, the probability of taking out orange flavoured candy is \(0.\)

(ii) As the bag has only lemon flavored candy, every time she will take out only lemon flavored candy. Therefore, the event is a sure event and the probability of taking out lemon flavored candy is \(1\).

Chapter 15 Ex.15.1 Question 7

It is given that in a group of \(3\) students, the probability of \(2\) students not having the same birthday is \(0.992.\) What is the probability that the \(2\) students have the same birthday?

 

Solution

Video Solution

 

What is known?

It is given that in a group of \(3\) students, the probability of \(2\)  students not having the same birthday is \(0.992.\)

What is unknown?

The probability that the \(2\) students have the same birthday.

Reasoning

This question is straight forward. We know that the sum of two complementary events are \(P\,(E) + P \) (not \(E\)\(=1.\)

By putting the given values in the above equation, we can find out the probability of not happening of event.

Steps:

Probability that \(2\) students not having the same birthday \( P\) (not \(E\)\(= 0.992\)

Probability that \(2\) students having the same birthday

\[\begin{align}P(E)&=1-0.992 \\& =0.008 \\\end{align}\]

Thus, the probability that the \(2\) students have the same birthday is \(0.008\)

Chapter 15 Ex.15.1 Question 8

A bag contains \(3\) red balls and \(5\) black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) red?     

(ii) not red?

 

Solution

Video Solution
 

 

What is known?

A bag contains \(3\) red balls and \(5\) black balls. A ball is drawn at random from the bag.

What is unknown?

The probability that the ball drawn is (i) red ? (ii) not red.

Reasoning:

This question can be solved easily in two steps;

(i) First find out the probability of drawing the red ball by using the formula:

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

(ii) Then by using the formula of sum of complementary event find out the probability of not getting a red ball i.e.

 \( \begin{align} P(\text{R})&=1-P(\text{ not E}) \\ & =1-\frac{3}{8} \\ & =\frac{5}{8} \end{align} \)

Steps:

No of red balls in a bag \(=3\)

No of black balls in a bag \(=5\)

\[\begin{align} \text{Total no of balls} &={3+5} \\ &=8 \end{align}\]

(i) Probability of drawing red ball =\(\begin{align}\frac{3}{8}\end{align}\)

(ii) Probability of not getting red ball 

\[\begin{align} P({R})&=1-{P}(\text { not } {E}) \\&=1-\frac{3}{8} \\ &=\frac{5}{8} \end{align}\]

Chapter 15 Ex.15.1 Question 9

A box contains \(5\) red marbles, \(8\) white marbles and \(4\) green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

(i) red?       

(ii) white?       

(iii) not green?

 

Solution

Video Solution
  

 

What is known?

A box contains \(5\) red marbles, \(8\) white marbles and \(4\) green marbles. One marble is taken out of the box at random.

What is unknown?

The probability that the marble taken out will be (i) red? (ii) white? (iii) not green?

Reasoning:

This question can be solved easily;

(i) Find out the probability of getting red, white and green marble by using the formula

Probability of an event 

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

(ii) Then by using the formula of sum of complementary event find out the probability of not getting green ball i.e.  \(P\,(E) +P\) (not \(E\)) = 1.

Steps:

No of red balls in a bag \(=5\)
No of white balls in a bag \(=8\)
No of green balls in a bag \(=4 \)
Total no of balls \(= 5+8+ 4 =17\)

Probability of drawing red ball\(\begin{align}=\frac{5}{17}\end{align}\)

Probability of drawing white ball\(\begin{align}=\frac{8}{17}\end{align}\)

Probability of drawing a ball which is green\(\begin{align}=\frac{4}{17}\end{align}\)

\[\begin{align} P(\text{not }E) & =1- P(E) \\ {} & =1-\frac{4}{17} \\ & =\frac{13}{17} \\\end{align}\]

Chapter 15 Ex.15.1 Question 10

A piggy bank contains hundred \(50\)\( \rm{p}\) coins, fifty \(\rm{Re}\) \(1\) coins, twenty \(\rm{Rs}\) \(2\) coins and ten \(\rm{Rs}\) \(5\) coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin

(i) will be a \(50\)\( \rm{p}\) coin?

(ii) will not be a \( \rm{Rs}\) \(5\) coin?

 

Solution

Video Solution
  

 

What is known?

A piggy bank contains hundred \(50\) \(\rm{p}\)  coins, fifty \( \rm{Re}\) \(1\) coins, twenty \(\rm{Rs}\) \(2\) coins and ten \(\rm{Rs}\) \(5\) coins. It is equally likely that one of the coins will fall out when the bank is turned upside down.

What is unknown?

The probability that the coin

(i) will be a \(50\)\( \rm{p}\) coin?

(ii) will not be a \( \rm{Rs}\) \(5\)  coin?

Reasoning:

This question can be solved easily;

Find out the probability of getting \(50\)\( \rm{p}\) coin, \( \rm{Re}\) \(1\) coin and \( \rm{Rs}\) \(2\) coin by using the formula

Probability of an event 

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

We can use same approach as we used in question no 9.

Total no of coins

\(=100+50+20+10=180\)

No of \(50\) \(\rm{p}\) \(=100\)
No of \(1 \) \( \rm{Re}\) coins \(=50\)
No of \(2\) \( \rm{Rs}\) coins \(=20\)
No of \(5\) \( \rm{Rs}\) coins\(=10\)

(i) Probability of drawing \(50\)\( \rm{p}\) coin \(=\frac{100}{180}\)

(ii) Probability of getting a \( \rm{Rs}\) \(5\) coin \(=\frac{10}{180}=\frac{1}{18} \)

Probability of not getting a \( \rm{Rs}\) \(5\) coin \(=1-\frac{1}{18}=\frac{17}{18}\)

Chapter 15 Ex.15.1 Question 11

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing \(5\) male fish and \(8\) female fish . What is the probability that the fish taken out is a male fish?

 

Solution

Video Solution
   

 

What is known?

A tank containing \(5\) male fish and \(8\) female fish

What is unknown?

The probability that the fish taken out is a male fish?

Rasoning:

This question can be solved easily by using the formula

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

No of male fish \(=5\)
No of female fish \(=8\)
Total no of fishes 

\(5+8=13\)

Probability that the fish taken out is a male fish \(\begin{align}=\frac{\text{No of male fish}}{\text{Total no of fishes}}\end{align}\)

Probability that the fish taken out is a male fish \(\begin{align}=\frac{5}{13}\end{align}\)

Chapter 15 Ex.15.1 Question 12

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers \(1, 2, 3, 4, 5, 6, 7, 8\) see Figure. and these are equally likely outcomes. What is the probability that it will point

(i) at \(8\)?

(ii) an odd number?

(iii) a number greater than \(2\)?

(iv) a number less than \(9\)?

  

Solution

Video Solution

 

What is known?

The arrow will come to rest pointing at one of the numbers \(1, 2, 3, 4, 5, 6, 7, 8\) and these are equally likely outcomes. Totally there are \(8\) outcomes.

What is unknown?

The probability that it will point:

(i) at \(8\)?

(ii) an odd number?

(iii) a number greater than \(2\)?

(iv) a number less than \(9\)?

Reasoning:

This question can be solved easily by using the formula

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

(i) Total possible outcomes \(= 8\)

Probability of getting \(8\) =

\(\begin{align}\frac{\text{Probability of getting }8}{\text{Total no of outcomes}}\end{align}\)

Probability of getting \(8\) \(\begin{align}= \frac{1}{8}\end{align}\)

(ii)Total no of odd numbers\(=1,3,5,7 = 4\)

Probability of getting odd number \[\begin{align}&=\frac{\text{Total no of odd number}}{\text{Total no of outcomes}} \\ &=\frac{4}{8} \\ &=\frac{1}{2} \\\end{align}\]

(iii) Numbers greater than \(2\) are \(3,4,5,6,7,8= 6\)

Probability of getting numbers greater than \(2\)

\[\begin{align}&=\frac{\text{Numbers greater than 2}}{\text{Total no of outcomes}} \\ &=\frac{6}{8} \\ &=\frac{3}{4} \\\end{align}\]

(iv) Numbers less than \(9\) are \(1,2, 3,4,5,6,7,8= 8\)

Probability of getting numbers less than \(9\)

\[\begin{align}&=\frac{\text{ Numbers greater than 9}}{\text{Total no of outcomes}} \\& =\frac{8}{8}=1\end{align}\]

Chapter 15 Ex.15.1 Question 13

A die is thrown once. Find the probability of getting

(i) a prime number;

(ii) a number lying between \(2\) and \(6\);

(iii) an odd number.

   

Solution

Video Solution

 

What is known?

A die is thrown once. So \(1, 2, 3, 4, 5, 6\) are equally likely outcomes.

Totally there are \(6\) outcomes.

What is unknown?

The probability of getting

(i) a prime number;

(ii) a number lying between \(2\) and \(6\);

(iii) an odd number.

Reasoning:

This question can be solved easily by using the formula

Probability of an event 

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

No of outcomes when you throw a die

\( = (1,2,3,4,5,6 )= 6\)

No of  prime numbers on dice are \(1,3\) and \(5 = 3\)

(i) Probability of getting a prime number

\[\begin{align} \text{ }\,\,\text{}&={}\frac{\text{Number of prime numbers}}{\text{total no of outcomes}} \\&=\frac{3}{6} \\ &=\frac{1}{2} \\\end{align}\]

(ii)Numbers lying between \(2\) and \(6\) are \(3,4,5 = 3\)

Probability of getting a number lying between \(2\) and \(6\)

\[\begin{align}&=\frac{ \begin{Bmatrix} \text{ Number  lying } \\ \text{between }  2  \\ \text{ and }6 \end{Bmatrix} }{\text{total no of outcomes}} \\ &=\frac{3}{6} \\\end{align}\]

(iii)Total number of odd numbers are \(1,3\) and \(5 = 3\)

Probability of getting a odd number 

\[\begin{align}& =\frac{\text{Number of odd numbers}}{\text{total no of outcomes}} \\ {} & =\frac{3}{6} \\ \end{align}\]

Chapter 15 Ex.15.1 Question 14

One card is drawn from a well-shuffled deck of \(52\) cards. Find the probability of getting

(i) a king of red colour 

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

 

Solution

Video Solution

 

What is known?    

One card is drawn from a well-shuffled deck of \(52\) cards.

What is unknown?

The probability of getting

(i) a king of red colour 

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

This question can be solved easily by using the formula

Reasoning:

Probability of an event 

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

Total number of cards from a well-shuffled deck = \(52\)

No of spade cards \(=13 \)

No of heart cards \(=13\)

No of diamond cards\(=13\)

No of club cards\(=13\)

Total number of kings \(= 4\)

Total number of queens\(= 4\)

Total number of jack cards \(= 4\)

No of face cards \(= 12\)

(i) Probability of getting a king of red colour

\[\begin{align} \text{  }&=\frac{\text{Number of red colour king}}{\text{Total no of outcomes}} \\ &=\frac{2}{52}=\frac{1}{26} \end{align}\]

(ii) Probability of getting a face card

\[\begin{align} \text{  } & =\frac{\text{ Number of face cards }}{\text{ Total no of outcomes }} \\ {} & =\frac{12}{52}=\frac{3}{13} \\\end{align}\]

(iii) Probability of getting a red face card

\[\begin{align}&=\frac{\text{ Number of red face cards}}{\text{Total no of outcomes}} \\ &=\frac{6}{52}=\frac{3}{26} \\ \end{align}\]

(iv) Probability of getting the jack of hearts

\[\begin{align}&{ = \frac{{{\text{ Number of jack of hearts}}}}{{{\text{total no of outcomes}}}}}\\{}&{= \frac{1}{{52}}}
\end{align}\]

(v) Probability of getting a spade card

\[\begin{align}&=\frac{\text{ Number of spade cards}}{\text{Total no of outcomes}} \\&=\frac{13}{52}\\&=\frac{1}{4} \\\end{align}\]

 (vi) Probability of getting the queen of diamonds 

\[\begin{align} &= \frac{ \begin{Bmatrix} \text{ Number of} \\ \text{possible}  \\ \text{ outcomes}  \end{Bmatrix}}{\begin{Bmatrix}\text { Total no of} \\ \text{favorable} \\ \text{outcomes} \end{Bmatrix}} \\ & =\frac{1}{52} \\ \end{align}\]

Chapter 15 Ex.15.1 Question 15

Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is

(a) an ace?

(b) a queen?

 

Solution

Video Solution
  

 

What is known?

Five cards Sthe ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards and one card is then picked up randomly.

What is unknown?

(i)The probability of getting the queen

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Reasoning:

This question can be solved easily by using the formula

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

Total no of cards\(=5\)

No of queen cards \(=1\)

(i) Probability that the card is the queen

\[\begin{align} &= \frac{ \begin{bmatrix} \text{ Number of} \\ \text{possible outcomes} \end{bmatrix}}{\begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix}} \\ &=\frac{1}{5} \end{align}\]

(ii) If the queen is drawn and put aside, then four cards are left the ten, jack, king and ace of diamonds

(a) Probability that the card an ace

\[\begin{align} &= \frac{ \begin{bmatrix} \text{ Number of} \\ \text{possible outcomes} \end{bmatrix}}{\begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix}} \\ &=\frac{1}{4} \end{align}\]

(b) Probability that the card is the queen

\[\begin{align} &= \frac{ \begin{bmatrix} \text{ Number of} \\ \text{possible outcomes} \end{bmatrix}}{\begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix}} \\ & =\frac{0}{4} \\ &=0 \end{align}\]

Chapter 15 Ex.15.1 Question 16

\(12\) defective pens are accidentally mixed with \(132\) good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

 

Solution

Video Solution
   

 

What is known?

\(12\) defective pens are accidentally mixed with \(132\) good ones.

Total number of outcomes = \(144\)

What is unknown?

The probability of getting a good pen when \(1\) pen is picked randomly from the lot.

Reasoning:

This question can be solved easily by using the formula

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

No of defective pens \(= 12\)

No of good pens \(= 132\)

Total no of pens

\(=12+132\\ =144\)

Probability that the pen taken out is a good one

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ &=\frac{132}{144} \\ &=\frac{11}{12} \\\end{align}\]

Thus, the probability that the pen taken out is a good \( = \frac{11}{12}\)

Chapter 15 Ex.15.1 Question 17

(i) A lot of \(20\) bulbs contain \(4\) defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

 

Solution

Video Solution
 

 

What is known?

A lot of \(20\) bulbs contain \(4\) defective ones. One bulb is drawn at random from the lot.

What is unknown?

The probability of getting a defective bulb when \(1\) bulb is drawn randomly from the lot.

Reasoning:

This question can be solved easily by using the formula

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

Total no of bulbs = \(20\)

No of defective pieces = \(4\)

Probability that this bulb is defective

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\&=\frac{4}{20} \\ &=\frac{1}{5} \\\end{align}\]

Remaining total number of bulbs\(=\text{ }20-1=19\)

Remaining total number of non-defective bulbs \(=16-1=15\)

Probability that this bulb is not defective

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\& =\frac{15}{19} \end{align}\]

Chapter 15 Ex.15.1 Question 18

A box contains \(90\) discs which are numbered from \(1\) to \(90\). If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by \(5\).

 

Solution

Video Solution
 

 

What is known?

Total number of discs = \(90\)

Total number of 2digit numbers between \(1\) to \(90\) = \(81\)

Total number of perfect square numbers between \(1\) to \(90\) are \(1,4,9,16,25,36,49,64,81= 9\)

Total numbers that are divisible by \(5\) are \(5 \),\(10\),\(15\),\(20\),\(25\),\(30\),\(35\),\(40\),\(45\),\(50\),\(55\),\(60\),\(65\),\(70\),\(75\),\(80\),\(85\),\(90\)\(=18\)

What is unknown?

The probability that the disc bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by \(5\).

Steps:

(i) Probability of getting a two digit number

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorabl outcomes} \end{bmatrix} } \\& =\frac{81}{90}\\&=\frac{9}{10} \end{align}\]

(ii) Probability of getting a perfect square number

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ & =\frac{9}{90}\\&=\frac{1}{10} \end{align}\]

(iii) Probability of getting a number divisible by \(5\)

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ & =\frac{18}{90} \\  & =\frac{1}{5} \end{align}\]

Chapter 15 Ex.15.1 Question 19

A child has a die whose six faces show the letters as given below:

\(A\) \(B\) \(C\) \(D\) \(E\) \(A\)

The die is thrown once. What is the probability of getting

(i) \(\rm A?\) 

(ii) \(\rm D?\)

 

Solution

Video Solution
 

 

What is known?

A die has six faces which show the letters as shown in the diagram and it is thrown once.

\(A\) \(B\) \(C\) \(D\) \(E\) \(A\)

What is unknown?

The probability of getting (i) \(\rm A?\) (ii) \(\rm D?\)

Steps:

Total number of outcomes = \(6\)

(i) Probability of getting \(\rm A\)

\[\begin{align}& =\frac{\text{Number of possible outcomes}}{\text{No of favourable outcomes}} \\& =\frac{2}{6} \\\end{align}\]

(ii) Probability of getting \(\rm D\)

\[\begin{align}& =\frac{\text{Number of possible outcomes}}{\text{No of favourable outcomes}} \\ & =\frac{1}{6} \\\end{align}\]

The probability of getting \(\rm A\) and \(\rm D\) is \(\begin{align}\frac{2}{6}\text{ and }\frac{1}{6}\end{align}\)

Chapter 15 Ex.15.1 Question 20

Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter \(1\rm{m}\)?

 

Solution

Video Solution
 

 

What is known?

A die is dropped at random on the rectangular region as shown in figure.

Length of rectangular region \(=3\,\rm{m}\)

Breadth of rectangular region \(=2\,\rm{m}\)

Diameter of the circle \(=1 \,\rm{m}\)

∴ Radius of the circle \(=0.5\,\rm{m}\)

What is unknown?

The probability that the die dropped at random will land inside the circle with diameter \(1\,\rm{m}\)?

Steps:

Area of rectangular region

\[\begin{align}\text{}& =L\,\times B \\ {} & =3\,\times 2 \\{} & =6{{\text{m}}^{2}} \\\end{align}\]

Diameter of circular region \(=1 \rm{m}\) \(\) 

Radius of circular region \(\begin{align}=\frac{1}{2}\rm{m}\end{align}\) 

Area of circular region

\[\begin{align} &= \pi r^{2} \\ & =\pi \,\times {{\left( \frac{1}{2} \right)}^{2}} \\ &=\frac{\pi }{4}\end{align}\]

Probability that it will land inside the circle

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\  & =\quad \frac{\text{Area of circular region}}{\text{Area of rectangular region}} \\  & =\quad \frac{\frac{\pi }{4}}{6} \\ & =\frac{\pi }{24} \end{align}\]

The probability that it will land inside the circle is \(\begin{align}\frac{\pi}{24}\end{align}\)

Chapter 15 Ex.15.1 Question 21

A lot consists of \(144\) ball pens of which \(20\) are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

 

Solution

Video Solution
  

 

What is known?

A lot consists of \(144\) ball pens of which \(20\) are defective and the others are good. Nuri will buy a pen if it is good but will not buy if it is defective.

What is unknown?

The probability that

(i) She will buy it?

(ii) She will not buy it?

Reasoning:

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

Total no of ball pens } \(= 144 \)

No of defective ball pens\(= 20\)

No of good ball pens \(=144-20 =124\)

(i) Probability that Nuri will buy the pen

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\&=\frac{124}{144} \\ &=\frac{31}{36}  \end{align}\]

(ii) Probability that Nuri will not buy the pen 

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\& =\frac{20}{144} \\& =\frac{5}{36}  \end{align}\]

Chapter 15 Ex.15.1 Question 22

Refer to Example 13.

(i) Complete the following table:

Event 'Sum of 2 dice' \(2\) \(3\) \(4\) \(5\) \(6\) \(7\) \(8\) \(9\) \(10\) \(11\) \(12\)
Probability \[\begin{align}\frac{1}{36}\end{align}\]           \[\begin{align}\frac{5}{36}\end{align}\]       \[\begin{align}\frac{1}{36}\end{align}\]

 

(ii) A student argues that 'there are \(11\) possible outcomes\( 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\) and \(12\). Therefore, each of them has a probability \(1/11\). Do you agree with this argument? Justify your answer.

 

Solution

Video Solution
  

 

What is known?

A dice is thrown twice means total no. of possible outcomes are \(36\)

What is unknown?

The probability that sum will be \(3, 4, 5, 6, 7, 9, 10, 11?\)

Reasoning:

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

No of possible outcomes to get the sum as \(2\)

\[=\left( 1,1 \right)\ \]

No of possible outcomes to get the sum as \(3\)

\[=\left( 2,1 \right),\left( 1,2 \right) \]

No of possible outcomes to get the sum as \(4\)

\[=\left( 2,2 \right),\left( 1,3 \right),\left( 3,1 \right) \]

No of possible outcomes to get the sum as \(5\)

\[=\left( 3,2 \right),\left( 2,3 \right),\left( 4,1 \right),\left( 1,4 \right) \]

No of possible outcomes to get the sum as \(6\)

\[=\left( 5,1 \right),\left( 1,5 \right),\left( 3,3 \right),\left( 4,2 \right),\left( 2,4 \right) \]

No of possible outcomes to get the sum as \(7\)

\[ \begin{align}  =& \left( 4,3 \right),\left( 3,4 \right),\left( 6,1 \right), \\ &  \left( 1,6 \right),\left( 5,2 \right),\left( 2,5 \right) \end{align}  \]

No of possible outcomes to get the sum as \(8\)

\[=\left( 4,4 \right),\left( 6,2 \right),\left( 2,6 \right),\left( 5,3 \right),\left( 3,5 \right) \]

No of possible outcomes to get the sum as \(9\)

\[=\left( 5,4 \right),\text{ }\left( 4,5 \right),\left( 6,3 \right),\left( 3,6 \right) \]

No of possible outcomes to get the sum as \(10\)

\[=\left( 5,5 \right),\left( 6,4 \right),\left( 4,6 \right) \]

No of possible outcomes to get the sum as \(11\)

\[=\left( 6,5 \right),\left( 5,6 \right)\]

No of possible outcomes to get the sum as \(12\)

\[= \left( 6,6 \right)\]

 

 Event 'Sum of 2 dice'

\(2\) \(3\) \(4\) \(5\) \(6\) \(7\) \(8\) \(9\) \(10 \) \(11\) \(12\)
Probability \[\begin{align}\frac{1}{36}\end{align}\] \[\begin{align}\frac{2}{36}\end{align}\] \[\begin{align}\frac{3}{36}\end{align}\] \[\begin{align}\frac{4}{36}\end{align}\] \[\begin{align}\frac{5}{36}\end{align}\] \[\begin{align}\frac{6}{36}\end{align}\] \[\begin{align}\frac{5}{36}\end{align}\] \[\begin{align}\frac{4}{36}\end{align}\] \[\begin{align}\frac{3}{36}\end{align}\] \[\begin{align}\frac{2}{36}\end{align}\] \[\begin{align}\frac{1}{36}\end{align}\]

(ii)Probability of each of them is not \(1/11\) as these are not equally likely.

This is demonstrated in the solution of (i)

Chapter 15 Ex.15.1 Question 23

A game consists of tossing a one rupee coin \(3\) times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

 

Solution

Video Solution
  

 

What is known?

A game consists of tossing a one rupee coin \(3\) times

 \(= \begin{Bmatrix} \text{HHH,TTT,HTH,HHT,} \\ \text{THH,THT,TTH,HTT} \end{Bmatrix} =8\)

Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise.

What is unknown?

The probability that Hanif will lose the game.

Reasoning:

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Probability of not happening an event is equal to 1 minus probability of happening an event.

Steps:

Probability that Hanif will win the game

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ & =\frac{2}{8}\\&=\frac{1}{4} \end{align}\]

Probability that Hanif will lose the game

\[\begin{align} & =1-\frac{1}{4} \\ & \,=\frac{3}{4} \\\end{align}\]

The probability that Hanif will lose the game is \(\frac{3}{4} \)

Chapter 15 Ex.15.1 Question 24

A die is thrown twice. What is the probability that

(i) \(5\) will not come up either time?

(ii) \(5\) will come up at least once?

[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

 

Solution

Video Solution
  

 

What is known?

A die is thrown twice.

What is unknown?

The probability that

(i) \(5\) will not come up either time?

(ii) \(5\) will come up at least once?

Reasoning:

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Probability of not happening an event is equal to \(1\) minus probability of happening an event.

Steps:

(i)  No of possible outcomes when \(5\) will come up either time

\[\begin{align} & = \begin{bmatrix} \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right), \\ \left( 5,5 \right), \left( 5,6 \right),\left( 1,5 \right),\left( 2,5 \right), \\ \left( 3,5 \right),\left( 4,5 \right),\left( 6,5 \right) \end{bmatrix} \\&=11\end{align}\]

probability that \(5\) will come up either time 

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible  outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ & =\frac{11}{36} \end{align}\]

probability that \(5\) will not come up either time 

\[\begin{align} & =1-\frac{11}{36} \\{} & =\frac{25}{36} \\\end{align}\]

(ii) No of possible outcomes when \(5\) will come up at-least once = \(11\)

probability that \(5\) will come up at-least once

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ & =\frac{11}{36} \end{align}\]

probability that \(5\) will not come up either time is \(\begin{align}\frac{25}{36}\end{align}\) and probability that \(5\) will come up at least once is \(\begin{align}\frac{11}{36}\end{align}\)

Chapter 15 Ex.15.1 Question 25

Which of the following arguments are correct and which are not correct?

Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes-two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(1/3\).

(ii) If a die is thrown, there are two possible outcomes of an odd number or an even number. Therefore, the probability of getting an odd number is \(1/2\)

 

Solution

Video Solution

 

Reasoning:

Probability of an event

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

Steps:

(i) Incorrect

If two coins are tossed simultaneously then ,

Total possible outcomes are

\(\text{(H,H), (T,T), (H,T), (T,H)} = 4\)

No of outcomes to get two heads \(=\text{(H,H)}=1 \)

No of outcomes to get two tails \(=\text{(T,T)}= 1\)

No of outcomes to one of each\(=\text{(H,T), (T,H) }= 2\)

Probability of getting two head

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\& =\frac{1}{4} \end{align}\]

Probability of getting two tails

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\& =\frac{1}{4} \end{align}\]

Probability of getting one of each

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\  & =\frac{2}{4}\\&=\frac{1}{2} \end{align}\]

It can be observed that,

Thus, the probability of each of the outcome is not \(\begin{align}\frac{1}{3}.\end{align}\)

(ii) correct

Total no of possible outcomes when a dice is thrown \(= \left( \text{1,2,3,4,5,6} \right)\)

No of possible outcomes to get odd number \(\left( \text{1,3,5} \right)=3 \)

No of possible outcomes to get even number \(\left( \text{2,4,6} \right)=3 \)

probability of getting odd number

\[\begin{align}& =\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } \\ & =\frac{3}{6}\\&=\frac{1}{2} \end{align}\]

Thus, the probability of getting an odd number is \(\begin{align}\frac{1}{2}.\end{align}\)

In the introduction, the concept of experimental or empirical probabilities is explained first. Following this, a theoretical approach to probability is presented and it is mentioned that theoretical probability is also known as classical probability. Further, the formula for theoretical probability is presented as the division of the number of outcomes favorable to an event E by the of all possible outcomes of the experiment. Also, it is observed that the sum of the probabilities of all the elementary events of an experiment is 1. Further to this, concepts like impossible events, sure or certain, and complementary events are explained.

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