Question

a class consists of 10 boys and 8 girls.3 students are selected at random.what is the probabiolity that the selected group has
1. all boys
2. all girls
3. 1 boy and 2 girls
4. at least1 girl
5. at most 1 girl?

Answer 1

plz ask if u not understand

Answer 2

Step-by-step explanation:

A class consists of 10 boys and 8 girls.3 students are selected at random.what is the Probability that the selected group has

1. all boys:

Total Students=18

To choose all 3 boys

$P(3B) = \frac{^{10}C_3}{^{18}C_3} \\ \\ = \frac{10!}{7! \: 3!} \times \frac{15! \: 3!}{18!} \\ \\ = \frac{10 \times9 \times 8}{18 \times 17 \times 16} \\ \\ P(3B)=\frac{5}{34}$

2) All girls

$P(3G) = \frac{^8C_3}{^{18}C_3} \\ \\ = \frac{8!}{5! \: 3!} \times \frac{15! \: 3!}{18!} \\ \\ = \frac{8\times7 \times 6}{18 \times 17 \times 16} \\ \\ P(3G) = \frac{7}{102}$

3. 1 boy and 2 girls

$P(1B2G) = \frac{^{10}C_1 \: ^8C_2}{^{18}C_3} \\ \\ = \frac{5 \times 7 }{ 17 \times 6} \\ \\ P(1B2G)= \frac{35}{102}$

4. at least1 girl

$P(At1G) =P(3G) + P(2G1B) + p(1G2B) \\ \\ = \frac{^8C_3}{^{18}C_3} + \frac{^8C_2 \times ^{10}C_1}{^{18}C_3} + \frac{^8C_1\: ^{10}C_2}{^{18}C_3} \\ \\ = \frac{7}{102} + \frac{35}{102} +\frac{15}{34} \\\\=\frac{87}{102}$

5. at most 1 girl?

$P(Am1G) = P(3B) + P(1G2B) \\ \\ = \frac{^{10}C_3}{^{18}C_3} + \frac{^8C_1\:^{10}C_2}{^{18}C_3} \\ \\ = \frac{5}{34}+ \frac{15}{34} \\\\P(Am1G)=\frac{10}{17}$

Hope it helps you.