Sample 2024-25 Class 12 Maths Q8 Solution: P(Ā/B̄) from P(Ā), P(B̄) and P(A ∩ B)

The question

For any two events $A$ and $B$ ⁠, if $P(\bar A) = \frac{1}{2}$ ⁠, $P(\bar B) = \frac{2}{3}$ and $P(A \cap B) = \frac{1}{4}$ ⁠, then $P(\bar A/\bar B)$ equals:

(A) $\frac{3}{8}$

(B) $\frac{8}{9}$

(C) $\frac{5}{8}$

(D) $\frac{1}{4}$

Exam answer

$P(A) = 1 - \frac{1}{2} = \frac{1}{2}$ and $P(B) = 1 - \frac{2}{3} = \frac{1}{3}$ ⁠.

Why?

An event and its complement have probabilities that add to 1:

P(A) = 1 - P(\bar A)

\begin{aligned} P(A) &= 1 - \frac12 = \frac22 - \frac12 = \frac12 \\[4pt] P(B) &= 1 - \frac23 = \frac33 - \frac23 = \frac13 \end{aligned}

\begin{aligned} P(A \cup B) &= \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \\ &= \frac{6 + 4 - 3}{12} \\ &= \frac{7}{12} \end{aligned}

Why?

The addition rule:

\begin{aligned} P(A \cup B) &= P(A) + P(B) \\ &\quad - P(A \cap B) \end{aligned}

The common denominator is 12: $\frac12 = \frac{6}{12}$ ⁠, $\frac13 = \frac{4}{12}$ and $\frac14 = \frac{3}{12}$ ⁠.

\begin{aligned} &\frac{6}{12} + \frac{4}{12} - \frac{3}{12} \\ &= \frac{6 + 4 - 3}{12} \\ &= \frac{10 - 3}{12} = \frac{7}{12} \end{aligned}

\begin{aligned} P(\bar A/\bar B) &= \frac{P(\bar A \cap \bar B)}{P(\bar B)} \\ &= \frac{1 - P(A \cup B)}{P(\bar B)} \\ &= \frac{1 - \frac{7}{12}}{\frac{2}{3}} \\ &= \frac{5}{12} \times \frac{3}{2} = \frac{5}{8} \end{aligned}

Why?

Conditional probability: $P(E/F)$ is $P(E \cap F)$ divided by $P(F)$ ⁠.

"Not $A$ and not $B$ " is the complement of " $A$ or $B$ " (De Morgan's law), so

P(\bar A \cap \bar B) = 1 - P(A \cup B)

Divide by the given $P(\bar B) = \frac{2}{3}$ ⁠, not by $P(B)$ ⁠. Dividing by $\frac23$ is multiplying by $\frac32$ ⁠.

\begin{aligned} 1 - \frac{7}{12} &= \frac{12}{12} - \frac{7}{12} = \frac{5}{12} \\[4pt] \frac{5}{12} \times \frac{3}{2} &= \frac{5 \times 3}{12 \times 2} = \frac{15}{24} \\ &= \frac{15 \div 3}{24 \div 3} = \frac{5}{8} \end{aligned}

Answer: (C) $\dfrac{5}{8}$

Where this comes from

Chapter 13, Probability: Conditional probability with complements.

Question type: Given numbers

How to spot it: click to see

Type 1. Given numbers: divide by the event after the bar

Spot it: the question gives $P(E)$ ⁠, $P(F)$ and $P(E\cap F)$ ⁠, and asks for $P(E|F)$ or $P(F|E)$ ⁠.

Here: The numbers are given, and $\bar B$ is after the bar, so divide by $P(\bar B) = \frac23$ ⁠. The top, $P(\bar A \cap \bar B)$ ⁠, is $1 - P(A \cup B) = \frac{5}{12}$ from the addition rule.

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Q8: P(Aˉ/Bˉ)P(\bar A/\bar B)P(Aˉ/Bˉ) from P(Aˉ)P(\bar A)P(Aˉ)⁠, P(Bˉ)P(\bar B)P(Bˉ) and P(A∩B)P(A \cap B)P(A∩B)

Q8: $P(\bar A/\bar B)$ from $P(\bar A)$ ⁠, $P(\bar B)$ and $P(A \cap B)$