Question

The probability that a student is not a swimmer is $\dfrac{1}{5}$. Find the probability that out of 5 students
1. At least four are swimmers
2. At most three are swimmers

Answer 1

We have the probability of one student being a swimmer. So we have to find the probability of two different cases as given in the question. We have to use the formulas of probability to solve this question.

Complete step by step solution:
According to the question we have to find the probability that out of 5 students at least four are swimmers and other at most three are swimmers.
Let X denote the number of students, out of 5 students, who are swimmers.
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.
$P(X)$ is the probability of event X
Out of 5 boys the probability of at least 4 being a swimmer is,
$\Rightarrow P(4) = {}^5{C_4} \times {(\dfrac{4}{5})^4} \times (\dfrac{1}{5})$ (Probability of four boys to be swimmer)
$\Rightarrow P(5) = {(\dfrac{4}{5})^5}$ (Probability of five boys to be swimmer)

Hence the probability that at least four are swimmer = $P(4) + P(5) =$ ${}^5{C_4} \times {(\dfrac{4}{5})^4} \times (\dfrac{1}{5}) + {(\dfrac{4}{5})^5}$
$\Rightarrow 0.73728$
Hence, this is our answer for part one
Now, out of 5 boys the probability of at most 3 being a swimmer is,
$\Rightarrow P(X \le 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

It represents the probability of at most three boys to be swimmers.
$\Rightarrow {(\dfrac{1}{5})^5} + {}^5{C_1}(\dfrac{4}{5}){(\dfrac{1}{5})^4}$ $+ {}^5{C_2}{(\dfrac{4}{5})^2}{(\dfrac{1}{5})^3}$ $+ {}^5{C_3}{(\dfrac{4}{5})^3}{(\dfrac{1}{5})^2}$

Note:
We can observe that we have to take cases to solve this question. At most means that any value less then it or equal to it and at least means any value more than it or equal to it. We just have to understand the language of the question to solve it.