Question

In a binomial distribution, mean is 3 and variance is $\dfrac{3}{2}$. Find the probability of at least 5 successes.

Answer 1

Hint: In order to solve this question, we should know about the formula of binomial distribution, that is, the mean is given by np, variance is given by npq, where n represents the number of trials, p represents the probability of success and q represents the probability of loss and binomial probability for x success out of n trials is given by $P\left( X=x \right){{=}^{n}}{{C}_{r}}{{p}^{x}}{{q}^{n-x}}$. By using these formulas, we can solve this question.

Complete step-by-step answer:
In this question, we have been given mean as 3 and variance as $\dfrac{3}{2}$ of a binomial distribution. And we have been asked to find the probability of at least 5 successes. To solve this question, we should know that for any binomial distribution, we can write mean = np and variance = npq. So, we can write,
np = 3 ……… (i)
$npq=\dfrac{3}{2}.........\left( ii \right)$
Now, we will divide equation (ii) by equation (i). So, we get,
$\begin{aligned} & \dfrac{npq}{np}=\dfrac{\dfrac{3}{2}}{3} \\\ & q=\dfrac{1}{2} \\\ \end{aligned}$
Now, we know that, p and q represents the probabilities of success and loss respectively. So, we can say that,
p + q = 1
$\begin{aligned} & p+\dfrac{1}{2}=1 \\\ & p=1-\dfrac{1}{2}=\dfrac{1}{2} \\\ \end{aligned}$
Now, we will put the value of p in equation (i). So, we get,
$\begin{aligned} & n\times \dfrac{1}{2}=3 \\\ & n=6 \\\ \end{aligned}$
Now, we have been asked to calculate the probability of at least 5 successes. So, the cases that we will include will be of 5 success and 6 success.
Now, we know that binomial probability for x success out of n trials, is given by,
$P\left( X=x \right){{=}^{n}}{{C}_{r}}{{p}^{x}}{{q}^{n-x}}$
So, we can say that the probability of 5 success will be given as,
$P\left( X=5 \right){{=}^{n}}{{C}_{5}}{{p}^{5}}{{q}^{n-5}}$
And the probability of 6 success will be given as,
$P\left( X=6 \right){{=}^{n}}{{C}_{6}}{{p}^{6}}{{q}^{n-6}}$
Therefore, we can write the probability for at least 5 success as,
Probability of at least 5 successes = P = P (X = 5) + P (X = 6)
Now, we will put the values of P (X = 5) and P (X = 6). So, we get,
$P{{=}^{n}}{{C}_{5}}{{p}^{5}}{{q}^{n-5}}{{+}^{n}}{{C}_{6}}{{p}^{6}}{{q}^{n-6}}$
Now, we will put that values of p, q and n, that are, $p=\dfrac{1}{2},q=\dfrac{1}{2}$ and n = 6. Therefore, we get,
$\begin{aligned} & P{{=}^{6}}{{C}_{5}}{{\left( \dfrac{1}{2} \right)}^{5}}{{\left( \dfrac{1}{2} \right)}^{6-5}}{{+}^{6}}{{C}_{6}}{{\left( \dfrac{1}{2} \right)}^{6}}{{\left( \dfrac{1}{2} \right)}^{6-6}} \\\ & P{{=}^{6}}{{C}_{5}}{{\left( \dfrac{1}{2} \right)}^{6}}{{+}^{6}}{{C}_{6}}{{\left( \dfrac{1}{2} \right)}^{6}} \\\ \end{aligned}$
Now, we know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So, we get,
$\begin{aligned} & P=\dfrac{6!}{5!\left( 6-5 \right)!}\times {{\left( \dfrac{1}{2} \right)}^{6}}+\dfrac{6!}{6!\left( 6-6 \right)!}{{\left( \dfrac{1}{2} \right)}^{6}} \\\ & P=\dfrac{6!}{5!}\times {{\left( \dfrac{1}{2} \right)}^{6}}+\dfrac{6!}{6!}{{\left( \dfrac{1}{2} \right)}^{6}} \\\ & P=6\times \dfrac{1}{{{2}^{6}}}+\dfrac{1}{{{2}^{6}}} \\\ & P=\dfrac{1}{{{2}^{6}}}\left( 6+1 \right) \\\ & P=\dfrac{7}{64} \\\ \end{aligned}$
Therefore, we can say that the probability of at least 5 successes is $\dfrac{7}{64}$, when the mean is 3 and the variance is $\dfrac{3}{2}$.

Note: While solving this question, we need to remember that the sum of the probabilities of success and failure is always 1, that is, p + q =1. Also, we need to remember that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and 0! = 1. The possible mistakes we can make in this question, is by not adding P (X = 6) with P (X = 5 ) which will give us the wrong answer, as taking only P (X = 5) means finding the probability of 5 successes and not at least 5 successes.