The probability that a class session is on a Friday is 1/3, and the probability that it is a warm, sunny day is 1/5 . In general, the probability that a particular student is absent is 1/10, but the probability that this student is absent given that it is a warm, sunny Friday is 1/6 . (a) What is the probability that on a day when the class meets, it is a warm, sunny Friday and the student in question is absent? (b) What is the probability that it is a warm, sunny Friday given that the student is absent? Solution P( F ) = 1/3, probability of friday P( S ) = 1/5, probability of sunny and warm P( A ) = 1/10 probability of absent P( A| F, S ) = 1 /6, probability that a student is absent given friday and sunny and warm. a) we need to find P( A, F, S ) = probability that its friday, sunny/warm and student is absent = P( A | F, S ) * P( F, S ) assuming that friday and sunny/warm are independent events = P( A | F, S ) * P( F ) * P( S ) = 1/6 * 1/3 * 1/5 = 1/90. b) P( S, F | A ) using bayes theorm, the above P is same as P( A | F,S ) / P( A ) = (1/90)/(1/10 ) = 1/9.

1. The probability that a class session is on a Friday is 1/3 ,and the probability that it is a warm,
sunny day is 1/5 . In general, the probability that a particular student is absent is 1/10 , but the
probability that this student is absent given that it is a warm, sunny Friday is 1/6 . (a) (4 points)
What is the probability that on a day when the class meets, it is a warm, sunny Friday and the
student in question is absent? (b) (4 points) What is the probability that it is a warm, sunny
Friday given that the student is absent?
Solution
P( F ) = 1/3, probability of friday
P( S ) = 1/5, probability of sunny and warm
P( A ) = 1/10 probability of absent
P( A| F, S ) = 1 /6, probability that a student is absent given friday and sunny and warm.
a) we need to find P( A, F, S ) = probability that its friday, sunny/warm and student is absent
= P( A | F, S ) * P( F, S )
assuming that friday and sunny/warm are independent events
= P( A | F, S ) * P( F ) * P( S )
= 1/6 * 1/3 * 1/5
= 1/90.
b)
P( S, F | A )
using bayes theorm, the above P is same as
P( A | F,S ) / P( A ) = (1/90)/(1/10 ) = 1/9