Board 2025 Class 12 Maths Q30 Solution: Absent students: p and the mean, or the probability of a distinction

The question

(a) The probability distribution for the number of students being absent in a class on a Saturday is as follows :

\begin{array}{|c|c|c|c|c|} \hline X & 0 & 2 & 4 & 5 \\ \hline P(X) & p & 2p & 3p & p \\ \hline \end{array}

Where X is the number of students absent.

(i) Calculate p. (1 mark)

(ii) Calculate the mean of the number of absent students on Saturday. (2 marks)

OR

(b) For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.

Exam answer

(a)

(i) Since $\sum P(X) = 1$ ⁠,

Why?

In a probability distribution, $X$ takes exactly one of its listed values. So the probabilities of all the values add up to $1$ ⁠.

p + 2p + 3p + p = 1 \Rightarrow 7p = 1

$\therefore\ p = \frac17$ ⁠.

Why?

Check: the probabilities are $\frac17, \frac27, \frac37, \frac17$ ⁠. Each lies between $0$ and $1$ ⁠, and they add up to $1$ ⁠.

(ii)

\begin{aligned} \text{Mean} &= \sum X\,P(X) \\ &= 0(p) + 2(2p) \\ &\quad + 4(3p) + 5(p) \\ &= 0 + 4p + 12p + 5p \\ &= 21p = 21\times\frac17 \end{aligned}

Why?

The mean (expected value) of a random variable is the sum of each value times its probability:

E(X) = \sum x_i\,p_i

Write the products in terms of $p$ first, add them, and put $p = \frac17$ only at the end.

\begin{aligned} 0 + 4p &= 4p \\ 4p + 12p &= 16p \\ 16p + 5p &= 21p \\[4pt] 21\times\frac17 &= \frac{21}{7} = 3 \end{aligned}

$\therefore$ Mean $= 3$ ⁠.

(b)

Let $E_1$ ⁠: the applicant is a male, $E_2$ ⁠: the applicant is a female, and $A$ ⁠: the candidate chosen has a distinction in the written test.

Why?

Every applicant is either a male or a female, and not both. So $E_1$ and $E_2$ are mutually exclusive and together cover every case. That is what the theorem of total probability needs.

P(E_1) = \frac13,\quad P(E_2) = \frac23

Why?

Two third of $3000$ is $2000$ females, so there are $1000$ males:

\begin{aligned} \frac23 \times 3000 &= \frac{6000}{3} = 2000 \\ 3000 - 2000 &= 1000 \\[4pt] P(E_1) &= \frac{1000}{3000} = \frac13 \\ P(E_2) &= \frac{2000}{3000} = \frac23 \end{aligned}

P(A|E_1) = 0.4,\quad P(A|E_2) = 0.35

Why?

"The probability of a male getting a distinction" is a conditional probability: it is the chance of $A$ given that the applicant is a male.

By the theorem of total probability,

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac13\times0.4 + \frac23\times0.35 \\ &= \frac{0.4 + 0.7}{3} = \frac{1.1}{3} \end{aligned}

Why?

A distinction happens either with a male or with a female, so add the two products. $\frac23\times0.35 = \frac{0.7}{3}$ ⁠.

\begin{aligned} \frac13\times0.4 &= \frac{0.4}{3} \\ \frac23\times0.35 &= \frac{2 \times 0.35}{3} = \frac{0.7}{3} \\ \frac{0.4}{3} + \frac{0.7}{3} &= \frac{0.4 + 0.7}{3} = \frac{1.1}{3} \end{aligned}

Finally $\frac{1.1}{3} = \frac{11}{30}$ (multiply top and bottom by $10$ ⁠).

\frac{1.1}{3} = \frac{1.1 \times 10}{3 \times 10} = \frac{11}{30}

$\therefore$ Required probability $= \dfrac{11}{30}$ ⁠.

Where this comes from

Chapter 13, Probability: Mean of a distribution; total probability.

Question type: (b) Two routes to one result

How to spot it: click to see

(b) Type 7. Two routes to one result: total probability

Spot it: the result can happen in two (or more) ways, such as a ball from bag I or from bag II, and you want its overall chance.

Here: A distinction can come from a male (chance $\frac13$ ⁠, then 0.4) or a female (chance $\frac23$ ⁠, then 0.35). Multiply along each route and add: $\frac{11}{30}$ ⁠.

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Q30: Absent students: ppp and the mean, or the probability of a distinction

Q30: Absent students: $p$ and the mean, or the probability of a distinction