Question

Let A and B be two events in a sample space S such that P(A) = 0.5, P(B) = 0.6, and P(A intersectionB) = 0.15. Find the probabilities below. Hint: (A intersectionBc) union (A intersectionB) = A.
(a) P(A|Bc)

(b) P(B|Ac)

Answer 1

Answer:

(a) $\frac{7}{8}$

(b) $\frac{9}{10}$

Step-by-step explanation:

Given,

P(A) = 0.5 ⇒ $P(A^c)=1-P(A) = 1 - 0.5 = 0.5$

P(B) = 0.6 ⇒ $P(B^c)=1-P(B) = 1 - 0.6 = 0.4$

P(A∩B) = 0.15

∵ $P(A\cap B^c)=P(A) - P(A\cap B) = 0.5 - 0.15 = 0.35$

Similarly,

$P(B\cap A^c)=P(B) - P(B\cap A) = 0.6 - 0.15 = 0.45$

Now,

(a) $P(\frac{A}{B^c})=\frac{P(A\cap B^c)}{P(B^c)}=\frac{0.35}{0.4}=\frac{35}{40}=\frac{7}{8}$

(b) $P(\frac{B}{A^c})=\frac{P(B\cap A^c)}{P(A^c)}=\frac{0.45}{0.5}=\frac{45}{50}=\frac{9}{10}$