A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement. The probability that exactly two of the three balls were red, the first ball being red, is

Question

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement. The probability that exactly two of the three balls were red, the first ball being red, is (A) (1/3) (B) (4/7) (C) (15/28) (D) (5/28)

Tardigrade Admin · Accepted Answer

Since, the first ball drawn is red so, we are left with

4

red and

3

blue balls. Now, we draw two balls out of which one should be red and other should be blue.

∴

Required probability

= P {(RB), (BR)}

= P (RB) + P (BR)

= \frac{4}{7} \cdot \frac{3}{6} + \frac{3}{7} \cdot \frac{4}{6}

= 2 \times \frac{2}{7} = \frac{4}{7}

Q. A bag contains $5$ red and $3$ blue balls. If $3$ balls are drawn at random without replacement. The probability that exactly two of the three balls were red, the first ball being red, is

$\frac{1}{3}$

$\frac{4}{7}$

$\frac{15}{28}$

$\frac{5}{28}$

Q. A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement. The probability that exactly two of the three balls were red, the first ball being red, is

31​

74​

2815​

285​

Since, the first ball drawn is red so, we are left with 4 red and 3 blue balls. Now, we draw two balls out of which one should be red and other should be blue. ∴ Required probability =P{(RB),(BR)} =P(RB)+P(BR) =74​⋅63​+73​⋅64​ =2×72​=74​

Q. A bag contains $5$ red and $3$ blue balls. If $3$ balls are drawn at random without replacement. The probability that exactly two of the three balls were red, the first ball being red, is

$\frac{1}{3}$

$\frac{4}{7}$

$\frac{15}{28}$

$\frac{5}{28}$