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wrote...
Posts: 236
2 weeks ago
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.
Textbook 

Stats: Modeling the World


Edition: 4th
Authors:
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Posts: 239
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2 weeks ago
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a.  P(neither ham nor turkey) = 1 - P(ham ∪ turkey)
                                                     = 1 - [P(ham) +  P(turkey) - P(ham ∩ turkey)]
                                            = 1 - [0.44 + 0.58 - 0.16] = 1 - 0.86 = 0.14
        Or, using the Venn diagram below, 14%

b. P(ham only) = P(ham) - P(ham ∩ turkey) = 0.44 - 0.16 = 0.28  
Or, using the Venn diagram above, 28%
c. P(ham ) = = = 0.2759
d. No, the events are not disjoint, since some families (16%) have both ham and turkey at their holiday meals.
This verified answer contains over 3500 words.
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