a family wants to have 3 kids, a) what is the probability they will have 1 boy? b)what is the probability they will have at least 2 boys? c)what is the probability that they will have 1 girl? d) what is the probability that the family will not have all the same sex children?
what is the probability of the family having a certain sex child?
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Even though it is not actually accurate in the 21st century, we generally consider:
P(having a boy)=1/2
P(having a girl)=1/2
a) P(1 boy)=P(Boy,Girl,Girl)+P(Girl,Boy,Girl)+P(Girl,Girl,Boy)=1/2*1/2*1/2+1/2*1/2*/2+1/2*1/2*1/2=1/8+1/8+1/8=3/8
b)P(at least two boys)=1-P(only girls)-P(1 boy)=1-P(Girl,Girl,Girl)-P(1 boy)=1-1/2*1/2*1/2-3/8=
1-1/8-3/8=1/2 (P(1 boy) has been calculated before)
c) P(1 girl)=P(1 boy)=3/8 (because the chances of having either a boy or a girl are the same)
d)P(all not the same sex)=1-P(all girls)-P(all boys)=1-P(girl, girl, girl)-P(boy,boy,boy)=1-1/2*1/2*1/2-1/2*1/2*1/2=1-2/8=1-1/4=3/4
I'm not sure what class you're in, and that's why I took a more simplistic approach. These probabilities can also easily be calculated with the TI83/84 by using the binompdf command.
The command has 3 inputs binompdf(n,p,x) so for calculating for example
c) P(having one girl)=binompdf(3,.5,1) - We want 3 children, the probability of having a girl in one try is 1/2=.5, and we wish to have 1 girl out of the three.
I encourage you to look up a youtube video for a more thorough explanation on how to use the calculator. (ti84 binompdf)
Keep in mind that in a certain scenario all the probabilities must add up to one, that's why I've often times chosen an easier route like: P(not having same sex)=1-P(having same sex children)
because: P(having same sex children)+P(not having same sex children) must = 1
Here p=.51 of having a boy so part one says that the probability is .51*.49*.49 (two girls). Part two talks about everything except no boys (.49^3) or one boy (part one) so subtract the two quantities from total of one (probabilities of all outcomes). Part c is similar to part a where you have .49*.51^2 (two boys and a girl) and part d is everything except three boys(.51^3) and three girls (.49^3) from a total of one (all probable outcomes)
Hope that helps,