Question

Pierre works five days a week. He has 12 shirts, 8 pants, 8 ties, and 4 jackets that he can wear to work. Of these, 4 shirts, 3 pants, 2 ties, and 2 jackets are blue. Each day he randomly selects one of each item to wear. Assume the selections are independent, and assume his butler launders the clothes every night so he has full closet each morning.

1. What is the probability pierres entire outfit next Monday will be blue?
2. What is the probability that Pierre will wear entirely blue outfits on Monday and Friday while wearing outfits which are not entirely blue on Tuesday through Thursday.
3. What is the probability that Pierre will wear entirely blue outfits on Monday through Friday while wearing outfits which are not entirely blue on Tuesday through Thursday?
4. What is probability Pierre will wear an entirely blue outfit on exactly 2 of the 5 days next week?

SHOW WORK

Answer 1

1.We need to find the probability of pierre choosing every piece of clothing which is blue under the assumption that his choices are independent.

therefore,

he can select a blue shirt from among 12 shirts out of which 4 are blue,in

shirt = C₄¹² or C(12,4) ways

he can select pants,ties,jacket respectively in,

pants = C(8,3) ways

ties = C(8,2) ways

jacket = C(4,2) ways

Since his selection of each piece of clothing is independent of each other,we can multiply these results to get the probability of choosing a blue outfit.

P(blue_outfit) = P(shirt) * P(pants) * P(tie) * P(jacket)

P(blue_outfit) = C(12,4) * C(8,3) * C(8,2) * C(4,2)

We must also note that probability of not wearing a completely blue outfit is 1 - P(blue_outfit)

2.This questions asks us about two conditions or situations i.e

1. probability of wearing blue outfit on monday and friday (notice the usage of and indicating independent events)

2. probability of wearing not entirely blue outfit from tuesday through thursday

We would first find the probability of the first condition:-

P(blue_monday_&_friday) = P(blue_outfit(monday)) * P(blue_outfit(friday))

since the probability of wearing a blue outfit is equal on any given day,therefore

P(blue_monday_&_friday) = P(blue_outfit)²

Probability of second condition:-

P(not_blue(tuesday through thursday)) = [1 - P(blue_outfit)] ³ (for tuesday,wednesday and thursday)

to get the final probability,we simply need to multiply these two probabilities

P(blue_outfit)² x [1 - P(blue_outfit)] ³

3.In the previous question,the days for wearing completely blue and not completely blue outfits were disjoint or non-overlapping.

However,this is not the situation in this question.

Pierre would be wearing entirely blue outfits from monday to friday(including friday) given by,

P(blue_outfit)⁵

or

Pierre would be wearing entirely only on monday and friday and not entirely blue from tuesday to thursday(including thursday) given by,

P(blue_outfit)² x [1-P(blue_outfit)]³

These two conditions are clearly alternatives,and therefore their respective probabilities should be added.

Therefore, the final solution is P(blue_outfit)^{5 +} P(blue_outfit)² x [1-P(blue_outfit)]³

4. In this question,Pierre can wear an entirely blue outfit on any of the 2 days from the 5 days.

Pierre can choose to wear an entirely blue suit on any of the 5 days of the week which can be represented as:-

P(blue_outfit) * 5

He can choose to wear another entirely blue outfit on any of the remaining 4 ways given by,

P(blue_outfit) * 4

To get the final answer,we need to multiply these probabilities as they are independent

Therefore,final answer is P(blue_outfit)² x 5 x 4