Sample 2025-26 Class 12 Maths Q31 Solution: Mehul and Rashi: chance of Rashi's selection and of at least one

The question

Two students Mehul and Rashi are seeking admission in a college. The probability that Mehul is selected is 0.4 and the probability of selection of exactly one of the them is 0.5. Chances of selection of them is independent of each other. Find the chances of selection of Rashi. Also find the probability of selection of at least one of them.

Exam answer

Let the events be E: Mehul is selected, and F: Rashi is selected.

E and F are independent, $P(E) = 0.4$ and

P(E \cap F') + P(F \cap E') = 0.5

Why?

"Exactly one is selected" happens in two ways: Mehul is selected and Rashi is not $(E \cap F')$ ⁠, or Rashi is selected and Mehul is not $(F \cap E')$ ⁠.

These two cannot happen together, so their probabilities add.

Let $P(F) = x$ ⁠. Then

P(E)P(F') + P(F)P(E') = 0.5

Why?

If E and F are independent, so are E and $F'$ ⁠, and F and $E'$ ⁠. So each "and" probability is a product:

P(E \cap F') = P(E)P(F')

Also $P(F') = 1 - x$ and $P(E') = 1 - 0.4 = 0.6$ ⁠, by the complement rule.

\begin{aligned} 0.4(1 - x) + x(1 - 0.4) &= 0.5 \\ 0.4 - 0.4x + 0.6x &= 0.5 \\ 0.2x = 0.1 \Rightarrow x &= 0.5 \end{aligned}

Why?

Expand the brackets and collect the $x$ terms: $-0.4x + 0.6x = 0.2x$ ⁠. Move 0.4 to the right: $0.5 - 0.4 = 0.1$ ⁠.

Then $x = \frac{0.1}{0.2} = 0.5$ ⁠.

\begin{aligned} 0.4(1 - x) &= 0.4 - 0.4x \\ x(1 - 0.4) &= 0.6x \\[4pt] 0.4 - 0.4x + 0.6x &= 0.5 \\ 0.4 + 0.2x &= 0.5 \\ 0.2x &= 0.5 - 0.4 = 0.1 \\ x &= \frac{0.1}{0.2} = \frac12 = 0.5 \end{aligned}

So, the probability of selection of Rashi $= 0.5$ ⁠.

Probability of selection of at least one of them:

\begin{aligned} &= 1 - P(E' \cap F') \\ &= 1 - P(E')P(F') \\ &= 1 - 0.6 \times 0.5 = 1 - 0.3 \end{aligned}

Why?

The opposite of "at least one is selected" is "neither is selected", that is $E' \cap F'$ ⁠. So use the complement rule.

$E'$ and $F'$ are independent too, with $P(E') = 0.6$ and $P(F') = 1 - 0.5 = 0.5$ ⁠.

Check another way, with $P(E \cap F) = 0.4 \times 0.5 = 0.2$ ⁠:

\begin{aligned} P(E \cup F) &= 0.4 + 0.5 - 0.2 \\ &= 0.9 - 0.2 \\ &= 0.7 \end{aligned}

And in the main working, $0.6 \times 0.5 = 0.3$ and $1 - 0.3 = 0.7$ ⁠.

$\therefore$ Probability of selection of at least one of them $= 0.7$ ⁠.

Where this comes from

Chapter 13, Probability: Independent events: exactly one, at least one.

Question type: At least one, or exactly one

How to spot it: click to see

Type 6. At least one, or exactly one: think "not" first

Spot it: independent events, with "at least one", "the problem is solved", "neither" or "exactly one".

Here: The selections are independent, and "exactly one" is 0.5: $0.4(1 - x) + 0.6x = 0.5$ gives $P(F) = 0.5$ ⁠. For "at least one", take 1 minus neither: $1 - 0.6 \times 0.5 = 0.7$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q31: Mehul and Rashi: chance of Rashi's selection and of at least one