Q.Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red, is
(A) $\dfrac{1}{3}$
(B) $\dfrac{4}{7}$
(C) $\dfrac{15}{28}$
(D) $\dfrac{5}{28}$

This refers to the box of $5$ red and $3$ black balls (Question 74), with three balls drawn one by one without replacement. Following the official reading, we want the probability of the ordered outcome red, red, black — the first ball red and exactly two of the three red: …

The setup (from Question 74)

The box has $5$ red and $3$ black balls ($8$ in all), and three balls are drawn one by one without replacement. We are told the first ball is red and asked for the probability that exactly two of the three drawn are red.

The computation

The first ball is red and exactly two of the three are red, taken as the ordered outcome red, red, black (R, R, B):

Multiplying along the branch:

$P=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{60}{336}=\frac{5}{28}.$ …

Method: Sequential draws via the multiplication theorem along a branch

Use this when balls are drawn one by one without replacement and you want the probability of a specific ordered sequence.

Steps

Step 1: Write the chain of conditional probabilities.

For a sequence of outcomes, multiply along the branch, updating counts after each draw:

Step 2: Update the pool at every step.

Each draw removes one ball, so both the favourable count and the total drop by one for the next factor. …

Common Mistakes

Mistake 1: Not updating the counts after each draw.

Why it's wrong: without replacement, after drawing a red the pool becomes $4$ red of $7$, etc.; reusing $5/8$ each time is the with-replacement error. Correct approach: multiply $\dfrac{5}{8}\cdot\dfrac{4}{7}\cdot\dfrac{3}{6}$, decrementing both counts.

Mistake 2: Summing every "exactly two red" order when the wording fixes one. …