Probability Question 198
Question: A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then the probability for the ball chosen be white is
[EAMCET 2003]
Options:
A) $ \frac{2}{15} $
B) $ \frac{7}{15} $
C) $ \frac{8}{15} $
D) $ \frac{14}{15} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let A be the event of selecting bag X, B be the event of selecting bag Y and E be the event of drawing a white ball,
then $ P(A)=1/2,P(B)=1/2 $ , $ P(E/A)=2/5 $
$ P(E/B)=4/6=2/3 $ . $ P(E)=P(A)P(E/A)+P(B)P(E/B)=\frac{1}{2}\cdot \frac{2}{5}+\frac{1}{2}\cdot \frac{2}{3}=\frac{8}{15} $ .
BETA
