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If the names of 5 boys and 7 girls are written on slips of paper in hat, what are the odds that the first two slips drawn from the hat will name 2 girls?
The odds are the ratio of an event occurring to that of its not occurring. The probability that the first slip drawn has a girl's name is 7/12. Having drawn a girl's name first, the probability of the second slip drawn having a girl's name is 6/11. Therefore \[P(2\ girl's\ names)=\frac{7\times6}{12\times11}=\frac{7}{22}\] The probability of the first two slips not naming two girls is given by \[P(other\ than\ 2\ girl's\ names)=1-P(2\ girl's\ names)=1-\frac{7}{22}=\frac{15}{22}\] Therefore the required odds are \[\frac{7}{22}:\frac{15}{22}=7:15\]