Question

The mean and variance of a binomial distribution are $4$ and $2$ respectively. Then the probability of $2$ successes is
1. $\dfrac{28}{256}$
2. $\dfrac{219}{256}$
3. $\dfrac{128}{256}$
4. $\dfrac{37}{256}$

Answer 1

Hint : To solve this above given question we can start solving this question by first finding out what is binomial distribution of probability. We will then use the concept we know that the mean of any binomial variable is np. Variance of a binomial variable is said to be npq. Here in both cases n stands for the total numbers of trials that take place in the event, p stands for the total number of probability of the trial being a success and q stands for the probability of the trial being a failure. Since a trial either has to be a failure or a success, we can say that the total number of trials will always be equal to the addition of p and q.

Complete step-by-step answer :
Now to briefly explain what binomial distribution of probability is we can say that is
$P(x=r){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$
Here r stands for success and n stands for number of trials taking place here.
Since we know that p stands for the probability of success and r for probability we can say that p and q when we take the sum of it will be equal to one. i.e.
$p+q=1$
Now as we know the mean and variance of binomial variable is
$Mean=np$; $Variance=npq$
Substituting the values we get
$np=4$ ; $npq=2$
Therefore
$4\times q=2$
$q=\dfrac{1}{2}$
Now we know that p will be $p=1-\dfrac{1}{2}$
$p=\dfrac{1}{2}$.
To find n substitute value of p in mean and we get
$n=4\times 2=8$
Now to get value of $P(x=2)$ substituting the values we get
$P(x=2){{=}^{8}}{{C}_{2}}{{p}^{2}}{{q}^{6}}$
This will give us
$P(x=2)=\dfrac{8!}{2!(6!)}\times {{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{1}{2} \right)}^{6}}$
Simplifying this
$P(x=2)=\dfrac{8.7}{2}\times {{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{1}{2} \right)}^{6}}$
$P(x=2)=28\times \dfrac{1}{{{2}^{8}}}$
$P(x=2)=\dfrac{28}{256}$
Hence the probability of getting 2 successes is $\dfrac{28}{256}$
So, the correct answer is “Option 1”.

Note: It isn’t possible that the p and q’s sum is not equal to one. In questions like this you can also be asked the probability of it being a loss in situations like that you can just subtract the answer we got here from one and get the probability of losses.