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 February 24th, 2014, 04:43 PM #1 Newbie   Joined: Feb 2014 Posts: 9 Thanks: 0 Probability of Choosing Boys & Girls I'm very confused on this particular test question and how my teacher got the answer. Was hoping someone could walk me through the problem. Refer to a family with 5 children; assume that it is equally probable for a boy to be born as it is for a girl to be born. What is the probability that there are at Least three girls given there are at least two boys? The answer is C(5,2)/26. I understand the sample space is 2^5 = 32, but I have no clue where he deducted the 6 from and I'm assuming the C(5,2) is because we stated we had at least 3 girls of the 5 so we need to choose 2 more spots? Any help or "Baby Steps" through this problem would be most appreciated.
 February 24th, 2014, 05:07 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,805 Thanks: 1045 Math Focus: Elementary mathematics and beyond Re: Probability of Choosing Boys & Girls $\frac{\binom{5}{2}}{\binom{5}{2}\,+\,\binom{5}{3}\ ,+\,\binom{5}{4}\,+\,\binom{5}{5}}\,=\,\frac{10}{2 6}\,=\,\frac{5}{13}$ Make sense now?
 February 24th, 2014, 07:29 PM #3 Newbie   Joined: Feb 2014 Posts: 9 Thanks: 0 Re: Probability of Choosing Boys & Girls So I take it that because we used the term "at least" we then add up the probabilities of each possible Sample Space is what I'm thinking?
 February 24th, 2014, 07:45 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,805 Thanks: 1045 Math Focus: Elementary mathematics and beyond Re: Probability of Choosing Boys & Girls Close. The sample space is the number of ways that the family can have at least two boys.
 February 24th, 2014, 09:17 PM #5 Newbie   Joined: Feb 2014 Posts: 9 Thanks: 0 Re: Probability of Choosing Boys & Girls Sorry, that's what I meant. I noticed there was no C(5,0) or C(5,1) for the sample space and "At least 2" could mean 2,3,4,5. Thank you for your clarification!
 February 25th, 2014, 08:34 PM #6 Global Moderator   Joined: Dec 2006 Posts: 18,954 Thanks: 1601 Given that there are at least two boys, the only way there can be at least three girls is for there to be exactly three girls and two boys. Hence the numerator is C(5,2) (or, equivalently, C(5,3)).

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