At start, we have:
- 7 boys
- 4 girls
It is given that the first selection was a girl. Since there were 4 girls, there is 3 left to be picked. So we have:
- 1 girl picked
- 7 boys to be picked
- 3 girls to be picked
We want the next 3 pickes to be boys.
The probability that the first pick will be a boy is the number of boys to be picked from over the total team left to be picked from. We have 7 boys and a total of 7 + 3 = 10 members, so:
[tex]P_1=\frac{7}{10}[/tex]Next, we want another pick of boy, but now we have got only
- 6 boys
- 3 girls
So, the probability of the second pick to be boy is:
[tex]P_2=\frac{6}{9}[/tex]And for the third, we have:
- 5 boys
- 3 girls
Probability of
[tex]P_3=\frac{5}{8}[/tex]Since we want these three to occur, the final probability is the product of them:
[tex]P=P_1\cdot P_2\cdot P_3=\frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}=\frac{7}{1}\cdot\frac{1}{3}\cdot\frac{1}{8}=\frac{7}{24}[/tex]So, the answer as a fraction in the lowest form is:
[tex]\frac{7}{24}[/tex]