Answer to Question #114650 in Statistics and Probability for Hanan Hamid

Question

Question 3:
A class contains 10 men and 20 women of which half men and half women have brown eyes. Find the probability that a person chosen at random is a men or brown eyed.

Question 4:
Let A and B are two events with P(AUB)=3/4, P(A ̅ )=2/3, P(A∩B)=1/4
Find: P(A), P(B), P(A∩B ̅)
[hint: (A ̅ ) represents the compliment of event A. according to compliment law of addition P(A ̅ )=1-P(A) also P(A∩B ̅ )=P(A)-P(A∩B)]

Answer 1

Question 3:

M=men,W=Women,B=person has brown eyes

$P(M)=\frac{10}{30}=\frac{1}{3}\\ P(F)=\frac{20}{30}=\frac{2}{3}\\ P(B)=\frac{15}{30}=\frac{1}{2}\\ P(B|M)=\frac{5}{10}=\frac{1}{2}\\ P(B|W)=\frac{10}{20}=\frac{1}{2}\\$

so, probability that a person chosen at random is a men or brown eyed is$P(M\cup B)$

$P(B|M)=\frac{P(M\cap B)}{P(M)}\\ P(M\cap B)=P(B|M)P(M)=\frac{1}{2}*\frac{1}{3}\\ P(M\cap B)=\frac{1}{6}\\$

and

$P(M \cup B)=P(M)+P(B)-P(M\cap B)\\ P(M \cup B)=\frac{1}{3}+\frac{1}{2}-\frac{1}{6}\\ P(M \cup B)=\frac{2}{3}=0.67\\$

probability that a person chosen at random is a men or brown eyed is 0.67

Question 4:

$P(A)=1-P(\bar{A})=1-\frac{2}{3}=\frac{1}{3}\\ \bold{P(A)=\frac{1}{3}}$

$P(A \cup B)=P(A)+P(B)-P(A\cap B)\\ \frac{3}{4}=\frac{1}{3}+P(B)-\frac{1}{4}\\ \bold{P(B)=\frac{2}{3}}$

$P(A\cap \bar{B})=P(A)-P(A\cap B)\\ P(A\cap \bar{B})=\frac{1}{3}-\frac{1}{4}\\ \bold{P(A\cap \bar{B})=\frac{1}{12}}\\$