let prob(female) = p
prob(male) = 5p
??? very unrealistic ???
p + 5p=1
p = 1/6 <---- prob(female)
so they could have a female as a first child, as a second child, as a third child, etc
expected outcome = 1/6 + (1/6)(5/6) + (1/6)(5/6)(5/6) + ....
this becomes the sum of an infinite series, with a = 1/6 and r = 5/6
= (1/6) / (1 - 5/6)
= (1/6)(6/1) = 1 ?????
There is of course a limit to the fertility of a mother.
The numbers are weird because your prob of male vs female is totally out of wack!!!
prob(5 children)
= must have had 4 boys, then a girl
= (5/6)^4 (1/6)
= 625/7776
at most 4 children
---> girl as first child
or ---> one boy, then girl
or ---> two boys, then girl
or ---> 3 boys, then girl
= 1/6 + 5/12 + 5/24 + 5/48
= 43/48
A couple will stop given birth immediately a
female child is born. If a male is 5 times more
likely to be born as a female. Find
a) The expected number of delivery of that
family.
b)the probability that they ll be 5 children in the family.
3) Probability that they ll be at most 4 children
in the family.
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