Question

The probability that a student is not a swimmer is $\frac{1}{5}$, Find the probability that out of $5$ students,
(i) at least four are swimmers and (ii) at most three are swimmers.

Answer 1

Let $X$ be the number of students out of $n=5$ students that is a swimmer.

It is a Bernoulli trial as they satisfy the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial..

P (not a swimmer) $q=\frac{1}{5}⟹p=1-q=\frac{4}{5}$

Since $X$ has a bionomial distribution, the probability of $x$ success in n-Bernoulli trials,

$P(X=x)={}^n{C}_x.p^x.q^{n–x}$, where $x=0,1,2,...,n$ and $(q=1–p)$

$\left(i\right)$

$P$(At least four students are swimmer),

$P(X\ge 4)=P(X=4)+P(X=5)$

$={}^5{C}_4{\left(\frac{4}{5}\right)}^4{\left(\frac{1}{5}\right)}^1$ $+{}^5{C}_5{\left(\frac{4}{5}\right)}^5{\left(\frac{1}{5}\right)}^0$

$P(X\ge 4)={}^5{C}_4\left(\frac{4^4}{5^5}\right)+\frac{4^5}{5^5}$

$=9\times \frac{4^4}{5^5}$
$=9\times \frac{256}{624\times 5}$
$=0.73728$

$\left(ii\right)$

$P$(At most three students are swimmer),

$P(X\le 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

Or
$P(X\le 3)=1-P(x\ge 4)$
$=1-0.73728$
$P(X\le 3)=0.26272$