Answer in Statistics and Probability for SKM #248131

A Mathematics professor is teaching both a morning and an afternoon section of introductory calculus. Let A={the professor gives a bad morning lecture} and B={the professor gives a bad afternoon lecture}. If P(A) = 0.3, P(B) = 0.2 and P(A ∩ B) = 0.1. Calculate the following probabilities:

(a) P(B|A)

(b) P(B’|A)

(c) P(B|A’ )

(d) P(B’|A’)

(e) If at the conclusion of the afternoon class, the professor is heard to mutter “what a rotten lecture”, what is the probability that the morning lecture was also bad?

Given that,

“p(A) = 0.3, p(B) = 0.2, p(A cap B) = 0.1”.

“a)”

“p(B|A)={P(Acap B)over p(A)}={0.1over0.3}={1over3}”

“b)”

“p(B’|A)=1-p(B|A)=1-{1over3}={2over3}”

“c)”

From part “b” above, we can obtain “p(Acap B’)” as follows,

“p(B’|A)={p(Acap B’)over p(A)}”

So,

“{2over3}={p(Acap B’)over0.3}implies p(Acap B’)={2over3}times0.3=0.2”

Now,

“P(B|A’ )={p(Acap B’)over p(A’)}={p(Acap B’)over1- p(A)}={0.2over 1-0.3}={0.2over 0.7}={2over7}”

“d)”

“P(B’|A’)=1-p(B|A’)=1-{2over7}={5over7}”

“e)”

We determine the conditional probability

“p(A|B)={p(Acap B)over p(B)}={0.1over 0.2}={1over2}”

Therefore, the probability that the morning lecture is bad given that the afternoon lecture is bad is “{1over2}.”