Question

Randomly two balls are drawn from a bag containing 3 green balls and 5 red balls. What is the probability that the balls drawn contain at least one red ball (without replacement)?

• A. $\frac{25}{28}$
• B. $\frac{5}{28}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A

$\frac{25}{28}$

Number of red balls = 5
Number of green balls = 3
Total number of balls = 8

First draw:
P (green ball) = $\frac{3}{8}$

Second draw:
P (green ball) = $\frac{2}{7}$

Therefore probability of getting a green ball in both the draws
= $\frac{3}{8}\times \frac{2}{7}=\frac{6}{56}=\frac{3}{28}$

Hence,
Probability of getting at least one red ball =
1 - Probability of getting all green balls.
$=1-\frac{3}{28}=\frac{25}{28}$