Question

Suppose that the probability of giving birth to a boy and the probability of giving birth to a girl are both 0.5. Find the probability that in a family of seve children, the following occurs. (a) All seven children are boys (b) There are two girls and five boys (c) The youngest child is a girl. (d) The oldest child is a girt (a) The probability is (Type an integer.or a simplified fraction)

Answer 1

a.0.0078125

b. 0.1640625

c. 0.0078125

d. 0.0078125

Explanation:

To solve this question we can use the binomial probability function:

`Prob(k\,successes\,in\,n\,trials)=\left(\begin{array}{c}n&k\end{array}\right) p^k q^((n-k))\\`

where

• n is the number of trials in this case the number of children
• k is the number of successes, we can think it as the number of boys
• p is the probability that is a boy
• q is the probability that is a girl,

In this case case p=q

Then

(a)
`Prob(B,B,B,B,B,B,B)=1*0.5^7`

(b)
`Prob(5\,boys\,in\,7\,children)=\left(\begin{array}{c}7&5\end{array}\right)0.5^7`

(c)
`Prob(G\,last\,and\,6\,boys)=Prob(B)^6Prob(G)=0.5^7`

(d)
`Prob(G\,first\,and\,6\,boys)=Prob(G)Prob(B)^6=0.5^7`

(a), (c) and (d) are equal as a result of equal probability of giving birth to a boy or a girl. (b) is different because it takes into account the different orders that births can occur