Test Index

CBSE Class 12 Math 2011 Solved Paper

© examsnet.com
Question : 29 of 29
Marks: +1, -0
Suppose 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male?
Assume that there are equal number of males and females.
Solution:  
Let the events M, F and G be defined as follows:
M: A male is selected
F: A female is selected
G: A person has grey hair
It is given that the number of males = the number of females
∴ P (M) = P (F) = 12\frac{1}{2}
Now, P (G/M) = Probability of selecting a grey haired person given that the person is a:
Male = 5% = 5100\frac{5}{100}
Similarly, P (G/F) = 0.25% = 0.25100\frac{0.25}{100}
A grey haired person is selected at random, the probability that this person is a male = P(M|G)
=
P(M)×P(G∣M)P(M)×P(G∣M)+P(F)×P(G∣F)\frac{P(M) \times P(G|M)}{P(M) \times P(G|M) + P(F) \times P(G|F)}
[Using Baye’s Theorem]
= 12×510012×5100+12×0.25100\frac{\frac{1}{2} \times \frac{5}{100}}{\frac{1}{2} \times \frac{5}{100} + \frac{1}{2} \times \frac{0.25}{100}}
= 51005100+0.25100\frac{\frac{5}{100}}{\frac{5}{100} + \frac{0.25}{100}}
= 55.25\frac{5}{5.25}
= 2021\frac{20}{21}
© examsnet.com
Go to Question:
Prev Question