Question

A die is tossed thrice. Success is getting $1$ or $6$ on a toss. The mean and the variance of number of successes
A. $\mu =1,{{\sigma }^{2}}=\dfrac{2}{3}$
B. $\mu =\dfrac{2}{3},{{\sigma }^{2}}=1$
C. $\mu =2,{{\sigma }^{2}}=\dfrac{2}{3}$
D. None of these

Answer 1

Hint : You must have proper knowledge about probability and binomial distribution to solve the question. Formula of probability , binomial distribution, mean and variance should be known to you. We will firstly find the probability of getting 1 or 6 then using the suitable formula we will find the mean and variance.

Complete step-by-step answer :
Probability means possibility. Basically probability is a measure of the likelihood of an event to occur. We can only predict the chance of an event to occur, that is how likely they are to happen . Probability can range in from 0 to 1 , where 0 means that the event is impossible and 1 indicates a certain event.
The probability of all the events in a sample space must be equal to 1.
Formula of probability is defined as the probability of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
That is P(E) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
Binomial distribution is the discrete probability that gives only two possible results of an experiment that is either success or failure.
There are two parameters n and p used in binomial distribution. The term ‘n’ states the number of times the experiment runs and the variable p tells the probability of any one outcome.
The binomial distribution formula is:
$P(x:n,p){{=}^{n}}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}$
Where,
n = the number of experiments
x= 0,1,2,3,…….
p = probability of success in a single experiment
q = probability of failure in a single experiment
For a binomial distribution, the mean and variance for the given number of success is represented using the formulas where mean is represented by $\mu$ and variance is represented by ${{\sigma }^{2}}$ .
$\mu =np$ and
${{\sigma }^{2}}=npq$
Now according to the question,
We are given that the number of trials are $3$ and the success is getting 1 or 6 .
So, $n=3$
And the probability(p) of getting 1 or 6 = $\dfrac{2}{6}$
That is p = $\dfrac{1}{3}$
And q = 1-p
That is $q=1-\dfrac{1}{3}$
$\Rightarrow$ q= $\dfrac{2}{3}$
So, mean( $\mu$ ) = $3\times \dfrac{1}{3}$
$\mu$ =1
And the variance( ${{\sigma }^{2}}$ )= $3\times \dfrac{1}{3}\times \dfrac{2}{3}$
That is ${{\sigma }^{2}}$ = $\dfrac{2}{3}$
So, the correct answer is “Option A”.

Note : A probability of 0 means that an event is impossible. A probability of 1 means that an event is certain. The probabilities of our different outcomes must sum to 1. In binomial probability , the number of observations n is fixed. Each observation is independent. Each observation represents success or failure.