Suppose a family has three children.find the probability that the first two children born are boys. What is the probability that the last two children are girls?

Question

Nathan Axel · Accepted Answer

#1/4 and 1/4#

There are 2 ways to work this out.

Method 1. If a family has 3 children, then the total number of different boy-girl combinations is 2 x 2 x 2 = 8

Of these, two start with (boy, boy ...) The 3rd child can be a boy or a girl, but it does not matter which. So, #P(B, B) = 2/8 = 1/4#

Method 2 . We can work out the probability of 2 children being boys as : #P(B, B) = P(B) xx P(B) = 1/2 xx 1/2 = 1/4#

In exactly the same way, the probability of the last two children both being girls can be: (B, G, G) or (G, G, G) #rArr # 2 of the 8 possibilities. So, #1/4#

OR: #P(? , G, G) = 1 xx 1/2 xx 1/2 = 1/4#

(Note: The probability of a boy or a girl is 1)

Jace Anderson · Answer

undefined The probability that the first two children born are boys is $ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} $.

Similarly, the probability that the last two children are girls is also $ \frac{1}{4} $.

Eva Adamson · Answer

undefined The probability that the first two children born are boys is $ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} $.

Similarly, the probability that the last two children are girls is also $ \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} $.