Question

The proportion of male and female students in a class is found to be 1 : 2. What is the probability that out of 4 students selected at random with replacement (i) 2 or more will
be females (ii) no male student is selected ?

Answer 1

Answer on Question #67801 – Math – Statistics and Probability

Question

The proportion of male and female students in a class is found to be $1:2$. What is the probability that out of 4 students selected at random with replacement

(i) 2 or more will be females

(ii) no male student is selected?

Solution

We do not know how many students in total there are in the class. Let $x$ be the number of male students. Then the number of female students in the class is $2x$ and total number of the students in the class is $3x$. Therefore, the probability for a randomly chosen student with replacement to be female is

$$p = \frac{2x}{3x} = \frac{2}{3}.$$

Number of students selected at random with replacement $n = 4$. We have Bernoulli trials process with $n = 4$ and $p = \frac{2}{3}$.

The probability of getting exactly $k$ successes in $n$ trials is given by

$$P(k) = \binom{n}{k} p^k (1 - p)^{n - k}.$$

(i) We have to find

$$P(2 \text{ or more females}) = P(2) + P(3) + P(4).$$

But it is easier to calculate probability for the complementary event:

$$\begin{array}{l} P(\text{less than 2 females}) = P(0) + P(1) = \binom{4}{0} \left(\frac{2}{3}\right)^0 \left(1 - \frac{2}{3}\right)^4 + \\ + \binom{4}{1} \left(\frac{2}{3}\right)^1 \left(1 - \frac{2}{3}\right)^3 = \frac{1}{81} + 4 \cdot \frac{2}{3} \cdot \frac{1}{27} = \frac{1}{81} + \frac{8}{81} = \frac{1}{9} \approx 0.11. \end{array}$$

Then

$$\begin{array}{l} P(2 \text{ or more females}) = 1 - P(\text{less than 2 females}) = 1 - \frac{1}{9} = \frac{8}{9} \approx \\ \approx 0.89. \end{array}$$

(ii) The event 'no male student is selected out of 4 students' is the same as '4 female students is selected out of 4 students'. The corresponding probability is

$$P(4) = \binom{4}{4} \left(\frac{2}{3}\right)^4 \left(1 - \frac{2}{3}\right)^0 = \frac{16}{81} \approx 0.20.$$

Answer: (i) $\frac{8}{9} \approx 0.89$; (ii) $\frac{16}{81} \approx 0.20$.

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