Quote:
A bag contains blue and red balls only.
If one ball is drawn, the probability of selecting a blue ball is 1/2.
If two balls are drawn randomly without replacement, the probability of selecting a blue and then a red ball is 5/18.
Two balls are drawn randomly without replacement, what is the probability of drawing a red ball, given that the first ball drawn was blue?
Step 1: Understanding the questionWhen probability of selecting a blue ball is 1/2 = 50%, probability of selecting a red ball is also 1/2 = 50%, therefore number of blue and red balls are equal.
Let number of blue balls = number of red balls = x; total number of balls = 2x
probability of selecting a blue = \(\frac{x}{2x}\) and then a red ball = \(\frac{x}{(2x-1) }\)
probability of selecting a blue and then a red ball = \(\frac{x}{2x}\) * \(\frac{x}{(2x-1)}\) = 5/18
Simplifying, x = 5
Number of blue balls = number of red balls = 5; Total number of balls = 10
Step 2: Calculationprobability of drawing a red ball, given that the first ball drawn was blue
When first blue ball is drawn, total number of balls left = 9
Probability of drawing a red ball out of 9 balls = 5/9
E is correct _________________
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