Question

1. a video game designer is building an algorithm that will randomly select your character’s default clothing – shirt and pants. the probability of your character getting a red shirt is 0.36. the probability of your character getting black pants is 0.44. the probability of your character getting a black shirt is 0.12. the probability of your character getting white pants is 0.08. these are all independent events. a. what is the probability that your character gets both a red shirt and black pants? b. what is the probability that your character gets both a red shirt and white pants? c. what is the probability that your character gets both a black shirt and white pants? d. what is the probability that your character gets either white pants or black pants?

Explanation:

Step1: Recall probability - multiplication rule for independent events

For two independent events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\).

Step2: Calculate probability for part a

Let \(A\) be the event of getting a red - shirt with \(P(A) = 0.36\), and \(B\) be the event of getting black pants with \(P(B)=0.44\). Then \(P(A\cap B)=0.36\times0.44 = 0.1584\).

Step3: Calculate probability for part b

Let \(A\) be the event of getting a red - shirt with \(P(A) = 0.36\), and \(B\) be the event of getting white pants with \(P(B)=0.08\). Then \(P(A\cap B)=0.36\times0.08 = 0.0288\).

Step4: Calculate probability for part c

Let \(A\) be the event of getting a black - shirt with \(P(A) = 0.12\), and \(B\) be the event of getting white pants with \(P(B)=0.08\). Then \(P(A\cap B)=0.12\times0.08 = 0.0096\).

Step5: Calculate probability for part d

For two mutually - exclusive events \(C\) (getting white pants) and \(D\) (getting black pants), \(P(C\cup D)=P(C)+P(D)\). Since \(P(C) = 0.08\) and \(P(D)=0.44\), then \(P(C\cup D)=0.08 + 0.44=0.52\).

Answer:

a. \(0.1584\)
b. \(0.0288\)
c. \(0.0096\)
d. \(0.52\)