Answer to Question #248131 in Statistics and Probability for SKM

Question

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?

Answer 1

Given that,

$p(A) = 0.3, p(B) = 0.2, p(A \cap B) = 0.1$.

$a)$

$p(B|A)={P(A\cap B)\over p(A)}={0.1\over0.3}={1\over3}$

$b)$

$p(B'|A)=1-p(B|A)=1-{1\over3}={2\over3}$

$c)$

From part $b$ above, we can obtain $p(A\cap B')$ as follows,

$p(B'|A)={p(A\cap B')\over p(A)}$

So,

${2\over3}={p(A\cap B')\over0.3}\implies p(A\cap B')={2\over3}\times0.3=0.2$

Now,

$P(B|A' )={p(A\cap B')\over p(A')}={p(A\cap B')\over1- p(A)}={0.2\over 1-0.3}={0.2\over 0.7}={2\over7}$

$d)$

$P(B'|A')=1-p(B|A')=1-{2\over7}={5\over7}$

$e)$

We determine the conditional probability

$p(A|B)={p(A\cap B)\over p(B)}={0.1\over 0.2}={1\over2}$

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