Probstat/week2 practice 1 solutions

This are solutions to Probstat/week2 practice 1.

Conditional probability review

1.1 There are 7 numbers divisible by 2 and 3 (0, 6, 12, 18, 24, 30, 36). There are 14 numbers divisible by 3. Let A be the event that a random number is divisible by 3. Let B be the event that a random number is divisible by 2. We want to compute P(B|A), which is P(AB)/P(A) = 7/14 = 1/2.

1.2 Using the definition of events Aand B from question 1.1, note that P(B) = 21/42 = 1/2. Thus P(B|A) = P(B). This implies that P(AB)=P(A)P(B). Therefore, both events are independent.

1.3 The number divisible by 4 and 3 are 0, 12, 24, and 36. Let C be the event that a random number is divisible by 4. Thus, P(C|A) = 4/14 = 1/7.

1.4 Note that P(C) = 11/42. This is not equal to P(C|A). Therefore both events are not independent.

2.1 P(C|S) = P(CS)/P(S) = 0.12 / (0.12+0.03) = 12/15 = 4/5.

2.2 P(C^c|S^c) = P(C^cS^c)/P(S^c) = 0.78 / (0.78+0.07) = 0.78/0.85. (Approximately 0.91764)

2.3 P(S|C) = P(SC)/P(C) = 0.12 / (0.12 + 0.07) = 0.12/0.19 (approximately 0.6316)

2.4 It predicts incorrectly when the person smokes and does not have cancer or when the person does not smoke but has cancer. The probability is P(SC^c U S^cC) = P(SC^c) + P(S^cC) = 0.07 + 0.03 = 0.1.