Question

A pair of dice is thrown $200$ times. If getting a sum of $9$ is considered as a success, find the variance of the number of successes.

Answer 1

Hint : This problem is based on the concept of probability. We have to find the variance of the number of successes. When we are given, getting a sum $9$ is considered a success.
We will first find the sample space and favourable outcome. then we find the probability of getting success . Probability is given by $P(E) = \dfrac{{no. \; of \; favourable \; event }}{{total \; no. \; of \;event}}$ . Then we will find the variance of success.

Complete step by step solution:
The given question is based on probability. Probability is the branch of mathematics that predicts how likely an event will happen. Probability of an event is denoted as P(E). It is always in between $0$ and $1$ .
The events which are likely to occur are called favourable events.
Let $E$ be a favourable event. Then the probability of the event $E$ is given as
$P(E) = \dfrac{{no. \; of \; favourable \; event }}{{total \; no. \; of \;event}}$ .
Consider the given question,
When a pair of dice is thrown, total no of events $=$ $36$ .
Let as assume the sample space for rolling a pair of dice are mentioned in the table is given below:

Dice 1↓	Dice 2 $\to$
1	2
1	$(1,1)$
2	$(2,1)$
3	$(3,1)$
4	$(4,1)$
5	$(5,1)$
6	$(6,1)$

From the above table, we get
Favourable outcome (getting a sum $9$ ) are $(3,6),(4,5),(5,4),(6,3)$
The total number of favourable outcome (getting a sum $9$ ) $=$ $4$
Let p denote the probability of getting a sum $9$ .
Then, $p = \dfrac{4}{{36}} = \dfrac{1}{9}$
And let ‘q denote the probability of not getting sum $9$ .
$\therefore q = 1 - \dfrac{1}{9} = \dfrac{8}{9}$ .
Let $n$ denote the number of time dice thrown. Hence $n = 200$
Then $variance = npq$
$variance$ $=$ $200 \times \dfrac{1}{9} \times \dfrac{8}{9} = 19.75$
Hence the variance is $19.75$ .
Hence, a pair of dice is thrown $200$ times. If getting a sum of $9$ is consider as success, the variance of the number of successes is $19.75$
So, the correct answer is “ $19.75$ ”.

Note : We note that
Probability always lies between $0$ and $1$ .
Variance is the measure of variability / spread in data from mean.
If $p$ denote the probability then , $mean = np$ and $variance = npq$