Suppose 25% of the population of a city read newspaper A, 20% read newspaper B,

Question

Suppose 25% of the population of a city read newspaper A, 20% read newspaper B, 13% read C, 10% read both A and B and 8% read both A and C and 5 % read B and C and 4% read all three. If a person is elected at random, what is the probability that he/she does not read any of the newspapers? In a certain city, 75% of the residents jog (J), 20% like ice cream (1) and 40% enjoy music (M). (a) Do you think that these numbers look right? 7. If 15% of the people jog and like ice cream, 30% jog and enjoy music, 10% like ice cream and music and 5% do all three, find the probability that (b) a resident will engage in at least one of the three activities (c) a resident engages in precisely one type of activity.

Explanation / Answer

6. P(A) = 25% = 0.25 P(B) = 20% = 0.2 P(C) = 13% = 0.13

P(A B) = 10% = 0.1 P(A C) = 8% = 0.08 P(B C) = 5% = 0.05

P(A B C) = 4% = 0.04

P(A U B U C) = P(A) + P(B) + P(C) - P(A B) - P(B C) - P(C A) + P(A B C)

= 0.25 + 0.2 + 0.13 - 0.1 - 0.08 - 0.05 + 0.04

= 0.39

=> P(A U B U C)c = 1 - 0.39 = 0.61

7. P(J) = 0.75 P(I) = 0.2 P(M) = 0.4

(a) The numbers look right.

(b) P(J I) = 15% = 0.15 P(J M) = 30% = 0.3 P(I M) = 10% = 0.1

P(J I M) = 5% = 0.05

P(J U I U M) = P(J) + P(I) + P(M) - P(J I) - P(J M) - P(I M) + P(J I M)

= 0.75 + 0.2 + 0.4 - 0.15 - 0.3 - 0.1 + 0.05

= 0.85

(c) P(J) + P(I) + P(M) - 2P(J I) - 2P(J M) - 2P(I M) + 3P(J I M)

= 0.75 + 0.2 + 0.4 - 2*0.15 - 2*0.3 - 2*0.1 + 3*0.05

= 0.4

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