NCERT Solutions For Class 10 Maths Chapter 15 Exercise 15.2

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Question 1

Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

(i) the same day? 

(ii) consecutive days? 

(iii) different days?

Solution

Video Solution
 

What is known?

Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day.

What is the unknown?

The probability that both will visit the shop on

(i) the same day?

(ii) consecutive days?

(iii) different days

Reasoning:

To solve this question, first find out the total number of outcomes and all the possible outcomes. Now, to find the probability use the formula given below

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 outcomes \(=\text{ }5\text{ }\times \text{ }5=25\)

(i) No of possible outcomes that both will visit the shop on the same day

\[\begin{align}&= \left( \text{T,T} \right)\text{,}\left( \text{W,W} \right), \left( \text{TH,TH} \right)\text{,}  \\ & \quad \;\; \left( \text{F,F} \right)\text{,}\left( \text{S,S} \right)\\&=5\end{align}\]

Probability that both will visit the shop on the same day 
\[\begin{align}&=\frac{\text{No of possible outcomes }}{\text{ Total no of outcomes}} \\ & =\frac{5}{25} \\\end{align}\] 

(ii) No of possible outcomes that both will visit the shop on consecutive days

\[\begin{align}&= \left( \text{t,w} \right)\text{,}\left( \text{w,th} \right)\text{,}\left( \text{th,f} \right)\text{,}\left( \text{f,s} \right)\text{,} \\ & \quad \left( \text{w,t} \right)\text{,}\left( \text{th,w} \right)\text{,}\left( \text{f,th} \right)\text{,}\left( \text{s,f} \right) \\&= 8\end{align}\]
Probability that both will visit the shop on the consecutive days
\[\begin{align}&=\frac{\text{No of possible outcomes}}{\text{Total no of outcomes}} \\ &=\frac{8}{25} \\\end{align}\]

(iii) No of possible outcomes that both will visit the shop on different days

\[\begin{align}&= \left( \text{t,w} \right)\text{,}\left( \text{w,th} \right)\text{,}\left( \text{th,f} \right)\text{,}\left( \text{f,s} \right)\text{,} \\ & \quad \left( \text{w,t} \right)\text{,}\left( \text{th,w} \right)\text{,}\left( \text{f,th} \right)\text{,}\left( \text{s,f} \right) \\&= 8\end{align}\]

Probability that both will visit the shop on the different days
\[\begin{align}&=1-\frac{5}{25} \\&=\frac{20}{25} \\&=\frac{4}{5}\\\end{align}\]

Question 2

A die is numbered in such a way that its faces show the numbers \(1, 2, 2, 3, 3, 6\). It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

What is the probability that the total score is (i) even? (ii) \(6\)? (iii) at least \(6\)?  

Solution

Video Solution
   

What is known?

A die is numbered in such a way that its faces show the numbers \(1, 2, 2, 3, 3, 6\). It is thrown two times and the total score in two throws is noted

What is  unknown?

What is the probability that the total score is

(i) even?

(ii) \(6\)?

(iii) at least \(6\)?  

Reasoning:

To solve this question, first find out the total number of outcomes and all the possible outcomes. Now, to find the probability use the formula given below

Probability

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

Step:

\(+\) \(1\) \(2\) \(2\) \(3\) \(3\) \(6\)
\(1\) \(2\) \(3\) \(3\) \(4\) \(4\) \(7\)
\(2\) \(3\) \(4\) \(4\) \(5\) \(5\) \(8\)
\(2\) \(3\) \(4\) \(4\) \(5\) \(5\) \(8\)
\(3\) \(4\) \(5\) \(5\) \(6\) \(6\) \(9\)
\(3\) \(4\) \(5\) \(5\) \(6\) \(6\) \(9\)
\(6\) \(7\) \(8\) \(8\) \(9\) \(9\) \(12\)

Total number of possible outcomes \(= 6 \times 6 =36\)

(i) No of possible outcomes when the sum is even \(= 18\)

 Probability that the total score is even
 \[\begin{align} & =\frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes }} \\& =\frac{18}{36}\\&=\frac{1}{2} \\\end{align}\]

(ii) No of possible outcomes when the sum is \(6 = 4\)

Probability that of getting the sum \(6\)
\[\begin{align} & =\frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes }} \\& =\frac{4}{36}\\&=\frac{1}{9}\end{align}\]

(iii) No of possible outcomes when the sum is at-least6(greater than \(5\)) \(= 15\)

Probability that of getting the sum at-least \(6 \)
\[\begin{align} & =\frac{\text{No of possible outcomes }}{\text{Total no of outcomes}} \\& =\frac{15}{36}\\&=\frac{5}{12}\end{align}\]

Question 3

A bag contains \(5\) red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Solution

Video Solution
 

What is known?

A bag contains \(5\) red balls and some blue balls and the probability of drawing a blue ball is double that of a red ball.

What is unknown?

The number of blue balls in the bag.

Reasoning:

First find out the total number of outcomes and all the possible outcomes. Now, to find the probability use the formula given below

Probability 

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

As it is given in the question that the probability of drawing a blue ball is double that of a red ball, that means

\(2 \times\)probability of drawing red ball = probability of drawing blue ball.

Substitute the values in the above and you can find out the number of blue balls.

Steps:

No of red balls \(= 5  \)
Let the no of blue balls be  \(x\)
Total number of balls \(= x+5\)

Probability of drawing red ball

\[\begin{align}&= \frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes }} \\&=\frac{5}{x+5}\end{align}\]

Probability of drawing blue ball

\[\begin{align}&= \frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes }} \\&=\frac{x}{x+5}\end{align}\]

Thus, the probability of drawing a blue ball is double that of a red ball

\[\begin{align}2\left( \frac{5}{x+5} \right)=\frac{x}{x+5}\end{align}\]

Hence the number of blue balls \(= 10\)

Question 4

A box contains \(12\) balls out of which \(x\) are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If \(6 \) more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find \(x\).

Solution

Video Solution
 

What is known?

A box contains  \(12 \)balls out of which are black and one ball is drawn at random from the box. If \( 6\) more black balls are put in the box, the probability of drawing a black ball is now double of what it was before.

What is unknown?

The probability that the ball drawn at random will be a black ball and the value of .

Reasoning:

First suppose the number of black balls as .Then find the total number of possible outcomes. Now, find the probability by using the formula

 Probability .

Now\( 6\) more balls are put in the box and the probability of drawing a black ball is now double of what it was before,

Now, the probability of drawing a black ball \(= 2 x\) probability of drawing black ball before.

Step:

Total Number of balls \(= 12\)

Let the number of black balls  \(= x\)

 Probability of getting black ball\[\begin{align} &= \frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes }} \\&=\frac{x}{12}\end{align}\]

If \(6\) more black balls are put in the box, the probability of drawing a black ball is now double of what it was before,

Total number of balls \(=\text{ }12+6\)

Number of black balls \(=\text{ }x+\text{ }6\)

Now,

\(2 \times\) Probability of drawing black ball before \(=\) probability of drawing black ball.

\(2 \times\) Probability of drawing black ball before \[\begin{align}=\frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes}}\end{align}\]

\[\begin{align} 2\left( \frac{x}{12} \right)&=\frac{x+6}{18} \\ 2x\,\,\times \,18&=12\left( x+6 \right) \\ 3x&=\,x+6 \\ 3x-x&=\,6 \\ 2x&=\,6 \\ x&=3 \end{align}\]

Number of black balls are \(3\)

Question 5

A jar contains \(24 \) marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(2/3\)· Find the number of blue balls in the jar.

Solution

Video Solution

What is known?

A jar contains \(24\) marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(2/3\).

What is the unknown?

The number of blue balls in the jar.

Reasoning:

It is given in the question that total number of balls are \(24\). suppose the number of green marbles be x. Now to find the number of blue balls, subtract the green balls from the total number of marbles i.e. number of blue balls are \(24 – x\). It is given in the question that probability of getting green ball is \(\begin{align}\frac{2}{3}\end{align}\) , probability of getting green ball \(\begin{align}=\frac{2}{3}\end{align}\)

Put the values in the above and you will get the value of \(x\) i.e. the number of blue balls.

Steps:

Total number of marbles = \(24\)

Let the green marbles be \(x\) and the blue colour marbles be \(24 -x\)

Probability of getting green marbles \( \begin{align}=\frac{2}{3}\end{align}\)

 \[\begin{align}\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} } &=\frac{2}{3} \\ \frac{x}{24}&=\frac{2}{3} \\ x& =\frac{2}{3}\times 24 \\ x& =16  \end{align}\]

Therefore, the number of green marbles = \(16\)

Hence, total number of blue marbles

\(\begin{align}   & =24-x \\ & =24-16=8 \\ \end{align}\)

Thus, the number of blue balls is \(8\).

Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 15 Exercise 15.2 for FREE
Ncert Class 10 Exercise 15.2
Ncert Solutions For Class 10 Maths Chapter 15 Exercise 15.2
  
Download SOLVED Practice Questions of NCERT Solutions For Class 10 Maths Chapter 15 Exercise 15.2 for FREE
Ncert Class 10 Exercise 15.2
Ncert Solutions For Class 10 Maths Chapter 15 Exercise 15.2
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