Question

Mark the correct alternative in each of the following: In a family of 3 children, the probability of having at least one boy is
(a)$\frac{7}{8}$
(b)$\frac{1}{8}$
(c)$\frac{5}{8}$
(d)$\frac{3}{4}$

Answer 1

SOLUTION :  

The correct option is (a) : ⅞.

Given : In a family of three children  

The possible outcomes are : BBB, BBG, BGB,GBB,BGG, GBG, GGB, GGG

Total number of possible outcomes = 8

Let E = Event of getting at least one boy  

Outcomes favourable to  E = BBB, BBG, BGB,GBB,BGG, GBG, GGB

No. of favorable outcomes = 7

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 7/8  

Hence, the required probability of getting at least one boy ,P(E) = 7/8

HOPE THIS ANSWER WILL HELP  YOU…

Answer 2

the correct option is (a).