A pair of dice is thrown 200 times. If getting a sum of 9 is considered a success, then find the mean and the variance respectively of the number of successes.

Question

A pair of dice is thrown 200 times. If getting a sum of 9 is considered a success, then find the mean and the variance respectively of the number of successes. (A) (400/9), (1600/81) (B) (1600/81), (400/9) (C) (1600/81), (200/9) (D) (200/9), (1600/81)

Tardigrade Admin · Accepted Answer

When a pair of dice is thrown, the sample space has

36

equally likely outcomes.
Outcomes favourable to sum of

9

are

(5, 4)

,

(4, 5)

,

(6, 3)

,

(3, 6)

only.
So

p =

probability (sum is

9

)

= \frac{4}{36} = \frac{1}{9}

,

∴ q = 1 - \frac{1}{9} = \frac{8}{9}

,

n = 200

Hence, mean

= n p = 200 (\frac{1}{9}) = \frac{200}{9}

and
variance

= n pq = 200 (\frac{1}{9}) (\frac{8}{9}) = \frac{1600}{81}

Q. A pair of dice is thrown $200$ times. If getting a sum of $9$ is considered a success, then find the mean and the variance respectively of the number of successes.

$\frac{400}{9}$ , $\frac{1600}{81}$

$\frac{1600}{81}$ , $\frac{400}{9}$

$\frac{1600}{81}$ , $\frac{200}{9}$

$\frac{200}{9}$ , $\frac{1600}{81}$

Q. A pair of dice is thrown 200 times. If getting a sum of 9 is considered a success, then find the mean and the variance respectively of the number of successes.

9400​, 811600​

811600​, 9400​

811600​, 9200​

9200​, 811600​

Q. A pair of dice is thrown $200$ times. If getting a sum of $9$ is considered a success, then find the mean and the variance respectively of the number of successes.

$\frac{400}{9}$ , $\frac{1600}{81}$

$\frac{1600}{81}$ , $\frac{400}{9}$

$\frac{1600}{81}$ , $\frac{200}{9}$

$\frac{200}{9}$ , $\frac{1600}{81}$