Mathematics, 07.03.2020 04:22 aboatright7410
Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case one of them does not show up. From past experience, Christine knows that there is a 55% chance that Dave will not show up and a 45% chance that Mike will not show up. Dave and Mike do not know each other and their decisions can be assumed to be independent.
a. What is the probability that both Dave and Mike will show up? (Round your answer to 2 decimal places.)
b. What is the probability that at least one of them will show up?
c. What is the probability that neither Dave nor Mike will show up? (Round your answer to 2 decimal places.)
Answers: 1
![P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25](/tpl/images/0537/6360/a1c8a.png)
![=1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75](/tpl/images/0537/6360/6dd23.png)
![P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25](/tpl/images/0537/6360/8a362.png)
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Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She...
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