Question

In the random experiment of tossing two unbiased dice, let $E$ be the event of getting the sum $8$ and $F$ be the event of getting even numbers on both the dice. Then$\left(\text{I}\right)P\left(E\right)=\frac{7}{36}\&\left(\text{II}\right)P\left(F\right)=\frac{1}{3}$ .Which of the following is the correct statement $?$

• A. Both $\left(\text{I}\right)$ and $\left(\text{II}\right)$ are correct
• B. Neither $\left(\text{I}\right)$ nor $\left(\text{II}\right)$ is correct
• C. $\left(\text{I}\right)$ is correct, $\left(\text{II}\right)$ is incorrect
• D. $\left(\text{I}\right)$ is incorrect, $\left(\text{II}\right)$ is correct

Answer 1

The correct option is B

Neither $\left(\text{I}\right)$ nor $\left(\text{II}\right)$ is correct

Explanation for the correct option:

Determine the correct statement:

The number of outcomes when two dices are thrown are as shown below:

$$\begin{gathered} \left(1,1\right)\left(1,2\right)\left(1,3\right)\left(1,4\right)\left(1,5\right)\left(1,6\right) \\ \left(2,1\right)\left(2,2\right)\left(2,3\right)\left(2,4\right)\left(2,5\right)\left(2,6\right) \\ \left(3,1\right)\left(3,2\right)\left(3,3\right)\left(3,4\right)\left(3,5\right)\left(3,6\right) \\ \left(4,1\right)\left(4,2\right)\left(4,3\right)\left(4,4\right)\left(4,5\right)\left(4,6\right) \\ \left(5,1\right)\left(5,2\right)\left(5,3\right)\left(5,4\right)\left(5,5\right)\left(5,6\right) \\ \left(6,1\right)\left(6,2\right)\left(6,3\right)\left(6,4\right)\left(6,5\right)\left(6,6\right) \end{gathered}$$

$E$ is the event of getting the sum $8$ .

$\Rightarrow E=\{\left(2,6\right),\left(3,5\right),\left(4,4\right),\left(5,3\right),\left(6,2\right)\}$

Then the number of outcomes when two dices are thrown gives the sum $8$ is$5$ then

$\left(\text{I}\right)P\left(E\right)=\frac{5}{36}$

$F$ is the event of getting even numbers on both the dice.

$\Rightarrow F=\{\left(2,2\right),\left(2,4\right),\left(2,6\right),\left(4,2\right),\left(4,4\right),\left(4,6\right),\left(6,2\right),\left(6,4\right),\left(6,6\right)\}$

The number of outcomes of the event of getting even numbers on both the dice is $9$ then

$\left(\text{II}\right)P\left(F\right)=\frac{9}{36}=\frac{1}{4}$

Thus, neither$\left(\text{I}\right)$ nor $\left(\text{II}\right)$ is correct.

Hence, option (B) is the correct answer.