CBSE Sample Paper 2022-23 · Section A · 1 mark

Q14: $P(A' \cap B')$ for independent events

The question

Given two independent events A and B such that P(A) =0.3, P(B) = 0.6 and $P(A' \cap B')$ is

(a) $0.9$

(b) $0.18$

(d) $0.1$

Exam answer

A and B are independent, so $A'$ and $B'$ are also independent.

Why?

Independence means the probability of both is the product:

P(A \cap B) = P(A)\,P(B)

A standard result says that if A and B are independent, so are their complements $A'$ and $B'$ ⁠. That lets us multiply $P(A')$ and $P(B')$ ⁠.

\begin{aligned} P(A' \cap B') &= P(A')\,P(B') \\ &= (1 - 0.3)(1 - 0.6) \\ &= 0.7 \times 0.4 = 0.28 \end{aligned}

Why?

The complement rule gives $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$ ⁠.

Check another way. By De Morgan's law $A' \cap B' = (A \cup B)'$ ⁠, and

\begin{aligned} &P(A \cap B) = (0.3)(0.6) = 0.18 \\[4pt] &P(A \cup B) \\ &= P(A) + P(B) - P(A \cap B) \\ &= 0.3 + 0.6 - 0.18 \\ &= 0.9 - 0.18 = 0.72 \end{aligned}

so $P(A' \cap B') = 1 - 0.72 = 0.28$ ⁠.

Watch out: $0.18 = 0.3 \times 0.6$ ⁠, option (b), is $P(A \cap B)$ ⁠, not $P(A' \cap B')$ ⁠.

Answer: (c) $0.28$

Where this comes from

Chapter 13, Probability: Independent events and their complements.

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: $A' \cap B'$ means neither happens. The events are independent, so multiply the chances of each failing: $(1 - 0.3)(1 - 0.6) = 0.28$ ⁠.

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Probability in past papers: question 29 of 38 · see them all

Q14: P(A′∩B′)P(A' \cap B')P(A′∩B′) for independent events