Q.If $P(A) = \dfrac{4}{5}$ and $P(A \cap B) = \dfrac{7}{10}$, then $P(B \mid A)$ is equal to
(A) $\dfrac{1}{10}$
(B) $\dfrac{1}{8}$
(C) $\dfrac{7}{8}$
(D) $\dfrac{17}{20}$

Concept: Conditional Probability — $P(B \mid A)$ is the probability of $B$ occurring given that $A$ has already occurred.

Step 1: Recall the formula for conditional probability:

Step 2: Substitute the given values:

$P(B \mid A) = \frac{\frac{7}{10}}{\frac{4}{5}}$ …

The core idea here is conditional probability — the chance that $B$ happens, given that we already know $A$ has happened. When you condition on $A$, you shrink the "universe" from the whole sample space to just the part where $A$ occurs. So $P(B \mid A)$ is not the raw probability of $B$; it's the proportion of $A$ that also lies inside $B$.

The formula that captures this is:

This works because $P(A)$ is the new "total" probability in the conditioned world, and $P(A \cap B)$ is the part of that world where $B$ also occurs.

Now let's apply it step by step.

1. Identify what's given.

   We have $P(A) = \frac{4}{5}$ and $P(A \cap B) = \frac{7}{10}$. Notice that $P(A \cap B)$ is already the probability that both $A$ and $B$ happen — exactly the numerator we need.

2. Write the conditional probability formula.

3. Substitute the numbers.

4. Simplify the fraction. Dividing by $\frac{4}{5}$ is the same as multiplying by its reciprocal $\frac{5}{4}$: $\frac{7}{10} \times \frac{5}{4} = \frac{7 \times 5}{10 \times 4} = \frac{35}{40}$ …

Method: Conditional probability by direct substitution

Use this whenever a problem gives you a joint probability $P(A\cap B)$ together with the probability of the conditioning event, and asks for one conditional probability.

Steps

Step 1: Identify which event is the condition (the "given").

In $P(\text{target}\mid \text{condition})$, the event after the bar is the condition. Its probability becomes the denominator — it is the new "whole world."

Step 2: Write the definition. …

Common Mistakes

Mistake 1: Dividing by the wrong marginal.

Why it's wrong: for $P(B\mid A)$ the conditioning event is $A$, so the denominator must be $P(A)$, not $P(B)$ (which isn't even given here). Correct approach: match the denominator to the event after the bar — $P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}$. …