ADDITIONAL PP 9.2
In this particular family, there are four children, all girls. The mother is pregnant again.
a. If chance alone is responsible for the gender of each child, what is the probability the next
child will be a girl?
b. The family now has five children. At the beginning of starting this family, before any
children are born, if chance alone is responsible for the gender of each child, what is the
probability they will all be girls?
c. It turned out that in this family, all five children were girls. Does it seem to you that chance
is a reasonable explanation of this result? Explain.
SOLUTION is on next page.
SOLUTION
a.
Let P = probability of the next child being a girl. If chance alone is responsible for the
gender of each child, then P = 0.50. Therefore,
p(the next child will be a girl) = 0.50
b.
Because there are five children, N = 5. Let P = the probability of a girl with each child. If
chance alone is responsible for the gender of each child, P = Q = 0.50. The number of P
events = 5. Entering Table B under N = 5, number of P events = 5 and P = 0.50, we find
p(exactly 5 girls) = 0.0312