Probability of at most 2 boys?

T

Timcago

Junior Member
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Apr 13, 2006
Messages
77
A couple wants to have seven children. What is the probability that they have at most 2 boys?

Can anyone help?
 
\(\displaystyle \L \sum\limits_{k = 0}^2 {\left( {\begin{array}{c}
7 \\
k \\
\end{array}} \right)\left( {\frac{1}{2}} \right)^7 }\)
 
uhh we haven't learned whatever that is yet.

I found an answer by listing all possible ways of not getting any boys (1 way ggggggg), 1 boy (7 ways), and 2 boys (21 ways?)

Then i found the total possible outcomes by doing 2*2*2*2*2*2*2 = 128

so 1/128 + 7/128 + 21/128 = 29/128 = 0.227

Is 0.227 the correct probability of getting at most 2 boys?
 
That's nice, but what's your plan when you get a question with bigger numbers?

What's the probability of getting at least 12 boys when randomly selecting a committee of 23 from a class of 74 people, assuming the class is evenly split between boys and girls?
 
Ya the method i am using now is only for now. We are not supposed to know the E thing yet.

I am sure when we do learn the E thing we will start tackling bigger problems like the one you listed.

Was my probability correct?
 
Hello, Timcago!

I found an answer by listing all possible ways of:
0 boys (1 way ggggggg), 1 boy (7 ways), and 2 boys (21 ways)

Then i found the total possible outcomes by doing: \(\displaystyle 2^7 \,=\,128\)

so: \(\displaystyle \:\frac{1}{128}\,+\,\frac{7}{128}\,+\,\frac{21}{128}\:=\:\frac{29}{128}\:\approx\:0.227\;\;\) Right!
Click to expand...

This is the proper method for "small" problems
. . except: did you really make those lists?

A little combinatory theory would have helped.

For 1 boy and 6 girls, there are: \(\displaystyle \:{7\choose1}\:=\:\frac{7!}{1!6!}\:=\:7\) ways.

For 2 boys and 5 girls, there are: \(\displaystyle \:{7\choose 2}\:=\:\frac{7!}{2!5!}\:=\:21\) ways.

 
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