How do I start this problem Example 2. A survey of families revealed that 58% of

Question

How do I start this problem

Example 2. A survey of families revealed that 58% of all families eat turkey at holiday meals,
44% eat ham, and 16% have both turkey and ham to eat at holiday meals.

(a) What is the probability that a family selected at random had neither turkey nor ham at
their holiday meal?
(b) What is the probability that a family selected at random had only ham without having turkey
at their holiday meal?
(c) What is the probability that a randomly selected family having turkey had ham at their
holiday meal?
(d) Are having turkey and having ham disjoint events? Explain.

Explanation / Answer

Probability that families eat turkey at holiday P(T) = 0.58

Probability that families eat ham at holiday P(H) = 0.44

Probability that families eat turkey and ham at holiday P(T and H) = 0.16

P(T or H) = P(T) + P(H) - P(T and H) = 0.58 + 0.44 - 0.16 = 0.86

(A)

the probability that a family selected at random had neither turkey nor ham at their holiday meal = 1 - P(T or H)

hence required probability = 1 - 0.86 = 0.14

(B)

Probability that family selected at random had only ham = P(H) - P(T and H) = 0.44 - 0.16 = 0.28

(C)

the probability that a randomly selected family having turkey had ham at their
holiday meal = P(T and H) = 0.16

(D)

having turkey and having ham disjoint events are not disjoint events as the intersection of these two events is not zero.

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