Board 2025 Class 12 Maths Q7 Solution: P(E/F̄) for independent events E and F

The question

If E and F are two independent events such that $P(E) = \frac{2}{3}$ ⁠, $P(F) = \frac{3}{7}$ ⁠, then $P(E/\overline{F})$ is equal to :

(A) $\frac{1}{6}$

(B) $\frac{1}{2}$

(C) $\frac{2}{3}$

(D) $\frac{7}{9}$

Exam answer

E and F are independent, so E and $\overline{F}$ are independent.

Why?

This is a standard result. E splits into the parts inside F and outside F:

\begin{aligned} P(E \cap \overline{F}) &= P(E) - P(E \cap F) \\ &= P(E) - P(E)P(F) \\ &= P(E)\,[1 - P(F)] \\ &= P(E)\,P(\overline{F}) \end{aligned}

\begin{aligned} P(E/\overline{F}) &= \frac{P(E \cap \overline{F})}{P(\overline{F})} \\ &= \frac{P(E)\,P(\overline{F})}{P(\overline{F})} \\ &= P(E) = \frac{2}{3} \end{aligned}

Why?

Conditional probability: the chance of E, given that $\overline{F}$ has happened, is

P(E/\overline{F}) = \frac{P(E \cap \overline{F})}{P(\overline{F})}

Then $P(\overline{F})$ cancels. In words: for independent events, knowing that F did not happen does not change the chance of E.

With numbers: $P(\overline{F}) = 1 - \frac{3}{7} = \frac{4}{7}$ ⁠.

\begin{aligned} P(E \cap \overline{F}) &= \frac{2}{3} \cdot \frac{4}{7} = \frac{8}{21} \\ P(E/\overline{F}) &= \frac{8}{21} \div \frac{4}{7} \\ &= \frac{8}{21} \cdot \frac{7}{4} = \frac{2}{3} \end{aligned}

Why?

\begin{aligned} P(\overline{F}) &= 1 - \frac{3}{7} = \frac{7}{7} - \frac{3}{7} = \frac{4}{7} \\[4pt] \frac{2}{3} \cdot \frac{4}{7} &= \frac{2 \times 4}{3 \times 7} = \frac{8}{21} \\[4pt] \frac{8}{21} \cdot \frac{7}{4} &= \frac{8 \times 7}{21 \times 4} \\ &= \frac{56}{84} \end{aligned}

To divide by a fraction, multiply by it turned upside down. Then cancel: divide top and bottom by 28.

\frac{56}{84} = \frac{56 \div 28}{84 \div 28} = \frac{2}{3}

Watch out: $\frac{8}{21}$ is $P(E \cap \overline{F})$ ⁠, not the answer. Dividing by $P(\overline{F})$ still has to be done.

Answer: (C) $\dfrac{2}{3}$

Where this comes from

Chapter 13, Probability: Independent events and conditional probability.

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: In $P(E/\overline{F})$ the event after the bar is $\overline{F}$ ⁠, so divide by $P(\overline{F}) = \frac47$ ⁠. Independence makes the top $P(E)\,P(\overline{F})$ ⁠, so the answer is just $P(E) = \frac23$ ⁠.

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Q7: P(E/F‾)P(E/\overline{F})P(E/F) for independent events E and F

Q7: $P(E/\overline{F})$ for independent events E and F