In a survey of 100 families, 55 read newspaper A, 45 read B and 40 read C; 25 read both A and B, 15 read both B and C, 20 read both A and C, and 10 read all three. One family is chosen at random. What is the probability that it reads exactly one of the three newspapers?

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Accepted Answer

1/2. Only A = 55 − 25 − 20 + 10 = 20, only B = 45 − 25 − 15 + 10 = 15 and only C = 40 − 20 − 15 + 10 = 15, a total of 50 families, so the probability is 50/100 = 1/2. For comparison 9/10 of the families read at least one paper and 3/10 read exactly two.

In a survey of 100 families, 55 read newspaper A, 45 read B and 40 read C; 25 read both A and B, 15 read both B and C, 20 read both A and C, and 10 read all three. One family is chosen at random. What is the probability that it reads exactly one of the three newspapers?

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