ดูต้นฉบับสำหรับ 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 '''A'''and '''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<sup>c</sup>|S<sup>c</sup>) = P(C<sup>c</sup>S<sup>c</sup>)/P(S<sup>c</sup>)''' = 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<sup>c</sup> U S<sup>c</sup>C) = P(SC<sup>c</sup>) + P(S<sup>c</sup>C)''' = 0.07 + 0.03 = 0.1.

==  Counting ==