Question

PLEASE HELP!

4. The last couple of years in the winter season, Mrs. Blair has had to take many days off for one of her kids being sick. She has determined that each kid has a 63% chance of getting sick at some point in the winter months. She has 4 kids.





a) What is the probability that all her kids get sick this winter?




b) What is the probability that at least one kid will get sick this winter?




5. Challenge Problem: If A U B = S (Sample Space), P (A and B) = .25, P (A) = .35, then P(B) = ???​

Answer 1

Explanation:

4a)

that means all 4 events happen in combination as one event.

so, we have to multiply the probabilities of all 4 events.

that means

0.63⁴ = 0.15752961... ≈ 0.16

so, there is a 16% chance that all 4 children get sick.

in other words, the probabilty for that is about 0.16.

b)

the probabilty of at least 1 kids getting sick is the opposite of the probability of no kids is getting sick.

the probability of getting sick is 0.63, so the probability of not getting sick is 1 - 0.63 = 0.37.

the probability of no kids getting sick is similar to a)

0.37⁴ = 0.01874161 ≈ 0.02

the probabilty of at least one kid getting sick is the opposite of that :

1 - 0.01874161... = 0.98125839 ≈ 0.98

5)

A u B is the sample space. so, any possible event is either part of A or B or both.

that means

P(A') = 0.35 = 1 - P(A)

P(A) = 1 - 0.35 = 0.65

P(A and B') = P(of every A that is not B) = 0.25.

because P(A) = 0.65

that means that

P(A and B) = P(every A that is also B) =

= 0.65 - 0.25 = 0.4

P(B) = P(A') + P(A and B) = 0.35 + 0.4 = 0.75

FYI

we could have gotten this directly from

P(A and B') = P(of every A that is not B = everything that is only A) = 0.25.

because

P(A and B')' = P(of everything that is not "only A" = everything that is B) = P(B) = 1 - 0.25 = 0.75