Question

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.)

Answer 1

Answer:

(a) 0.25

(b) 0.75

(c) 0.25

Step-by-step explanation:

We are given the Probability that Dave will not show up, P(D) = 0.55

So, Probability that Dave will show up, P(D') = 1 - P(D) = 1 - 0.55 = 0.45

Also, Probability that Mike will not show up, P(M) = 0.45

So, Probability that Mike will show up, P(M') = 1 - P(M) = 1 - 0.45 = 0.55

(a) Probability that both Dave and Mike will show up, P(D' $\bigcap$ M') = Probability that Dave will show up * Probability that Mike will show up

    = P(D') * P(M') = 0.45 * 0.55 = 0.25

(b) Probability that at least one of them will show up = P(D' $\bigcup$ M')

    P(D' $\bigcup$ M')  =  P(D') + P(M') - P(D' $\bigcap$ M')

                      = 0.45 + 0.55 - 0.25 = 0.75

(c) Probability that neither Dave nor Mike will show up = 1 - Probability that at least one of them will show up

    = 1 - P(D' $\bigcup$ M') = 1 - 0.75 = 0.25

Answer 2

Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

D = Dave will show up.

M =  Mike will show up.

Given:

$P(D^{c})=0.55\\P(M^{c})=0.45$

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

$P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25$

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

                                                   $=1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75$

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

$P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25$

Thus, the probability that neither Dave nor Mike will show up is 0.25.