Probability Question 325
Question: The probability that a man will be alive in 20 years is $ \frac{3}{5} $ and the probability that his wife will be alive in 20 years is $ \frac{2}{3} $ . Then the probability that at least one will be alive in 20 years, is
[Bihar CEE 1994]
Options:
A) $ \frac{13}{15} $
B) $ \frac{7}{15} $
C) $ \frac{4}{15} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ A $ be the event that the husband will be alive 20 years. $ B $ be the event that the wife will be alive 20 years.
Clearly $ A $ and $ B $ are independent events $ P(A\cap B)=P(A).P(B) $ Given $ P(A)=\frac{3}{5}, $
$ P(B)=\frac{2}{3} $
The probability that at least of them will be alive 20 years is
$ P(A\cup B)=P(A)+P(B)-P(A\cap B) $
$ =P(A)+P(B)-P(A)P(B)=\frac{3}{5}+\frac{2}{3}-\frac{3}{5}.\frac{2}{3}=\frac{9+10-6}{15}=\frac{13}{15} $
A liter : Required probability is $ 1-P(A $ and $ B $ both will die) $ =1-\frac{2}{5}\times \frac{1}{3}=1-\frac{2}{15}=\frac{13}{15}. $
BETA
