Question 211123: A couple plan to have exactly four children.
(a) Construct a tree diagram and list the sample space.
(b) Find the probability that the family has at least three boys.
show the work so i can check myself....thanks
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! a)
Let's build the tree:
Using the tree, we get the sample space:
{b,b,b,b}
{b,b,b,g}
{b,b,g,b}
{b,b,g,g}
{b,g,b,b}
{b,g,b,g}
{b,g,g,b}
{b,g,g,g}
{g,b,b,b}
{g,b,b,g}
{g,b,g,b}
{g,b,g,g}
{g,g,b,b}
{g,g,b,g}
{g,g,g,b}
{g,g,g,g}
So for instance {b,g,g,b} means that the couple had a boy, girl, girl, and then a boy (in that order).
Note: recall, the sample space is the set of ALL possible outcomes.
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b)
Since we want to know the chances of the couple having "at least three boys", this means they want to know the chances of having 3 boys OR 4 boys (since at least means that figure or more).
Looking back at the list of all possible outcomes (ie the sample space) from part a), we see that we have the combinations for 3 boys:
{b,b,b,g}, {b,b,g,b}, {b,g,b,b}, and {g,b,b,b}
So there are 4 cases where the couple would have 3 boys.
So =\frac{\text{Number of Cases with 3 boys only}}{\text{Total number in sample space}}=\frac{4}{16}=\frac{1}{4})
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Also, since there is only ONE way to have 4 boys (of a total of 4 children), this means that
=\frac{\text{Number of Cases with 4 boys only}}{\text{Total number in sample space}}=\frac{1}{16})
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Now simply add the two probabilities to find the chances of either one occurring:
=P(\text{3 boys OR 4 boys}))
=P(\text{3 boys ONLY})+P(\text{4 boys ONLY}))
=\frac{\text{Number of Cases with 3 boys only}}{\text{Total number in sample space}}+\frac{\text{Number of Cases with 4 boys only}}{\text{Total number in sample space}})
=\frac{4}{16}+\frac{1}{16})
=\frac{4+1}{16})
=\frac{5}{16})
So the probability of the couple having AT LEAST 3 boys is  which is 0.3125 in decimal form which gives a 31.25% chance.
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