CMA Foundation · Fundamentals of Business Mathematics and Statistics · Probability
An urn contains 5 red and 3 white balls. Two balls are drawn without replacement. Given that the second ball is red, what is the probability that the first ball was also red?
The probability is 4/7. The chance both are red is 20/56, and the chance the second is red is 35/56, which includes the white-first case. Dividing 20/56 by 35/56 gives 4/7 by the conditional probability formula.
- A5/7
- B4/7Correct
- C5/8
- D4/5
Explanation
P(R1 and R2) = 5/8 x 4/7 = 20/56. P(R2) = P(R1R2) + P(W1R2) = 20/56 + (3/8 x 5/7 = 15/56) = 35/56 = 5/8. So P(R1|R2) = 20/35 = 4/7. Option 4/5 is wrong as it ignores the white-first case in the denominator.
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