# Ex.15.1 Q25 Probability Solution - NCERT Maths Class 10

## Question

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 \(1/2\)

## Text Solution

**Reasoning:**

\[\text{Probability of an event}=\frac{\text{ Number of possible outcomes }}{\text{Total no of favourable outcomes}}\]

**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

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

Probability of getting two tails

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

Probability of getting one of each

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

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

\[\begin{align} & =\frac{\text{ Number of possible outcomes }}{\text{No of favourable outcomes}} \\ {} & =\frac{3}{6}\\&=\frac{1}{2} \\\end{align}\]

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