Probability question - arranging 20 pupils in a row - 8 boys and 12 girls

Question

We have 20 pupils in class, 12 girls and 8 boys. We arrange the pupils in a row, and now need to calculate the following probability:

a. The probability that Jana, one of the girls, will not stand next to another girl.

b. 4 pupils out of the 20 (12 girls and 8 boys) are randomly chosen for the class commitee, What is the probability that both genders will have representation in the commitee.

I tried many times with no success.

Thanks in advance!

Answer 1

For (a) there are two cases: Jana is at one end, or she isn't.

If she's at the end, pick which end ($2$), then pick which of the eight boys stands next to her ($8$). Then place the other $18$ children ($18!$).

If she's not at the end, pick where she stands ($18$), then pick the boy that stands on her left ($8$) and then on her right ($7$). Then place the other $17$ children ($17!$).

Hence the probability is

$$P_a = \frac{2 \cdot 8 \cdot 18! + 18 \cdot 8 \cdot 7 \cdot 17!}{20!} = \frac{18}{95}.$$

For (b) there are three cases:

• three girls, one boy;
• three boys, one girl;
• two of each.

So in order:

• Pick the three girls ($_{12}C_3$) and the boy ($8$);
• Pick the three boys ($_{8}C_3$) and the girl ($12$);
• Pick the two boys ($_{8}C_2$) and the two girls ($_{12}C_2$).

Hence:

$$P_b = \frac{8 (_{12}C_3) + 12 (_{8}C_3) + (_{12}C_2) (_{8}C_2)}{_{20}C_4}.$$

Answer 2

a) Permute two boys on either side of Jana: $\frac{8!}{6!}$. Treat this trio as a single object. So now we permute $18!$ elements giving us: $\frac{8!}{6!} \cdot 18!$. We now have to consider cases where Jana is at the front or end of the line. If she is at the front, we select one boy, which can be done in $8$ ways. We then permute the remaining $18!$ members. Multiply this by $2$ for symmetry at the end of the line.

So we have two disjoint cases, and we add them:

$$ \frac { 18! \cdot (\frac{8!}{6!} + 16) }{ 20!}$$

b) Consider the probability that only boys will have representation. This happens in $\binom{8}{4}$ ways. Similarly, an all girls committee happens in $\binom{12}{4}$ ways. By rule of sum, we add these cases. There are $\binom{20}{4}$ ways to choose the committee. So by inclusion-exclusion, we get:

$$1 - \frac{\binom{8}{4} + \binom{12}{4}}{\binom{20}{4}}$$

Answer 3

a. There are $20!$ ways to arrange the $20$ pupils. If Jana is at the beginning, there are $8$ possibilities for the boy at her side, and $18!$ for the rest. This makes $8\cdot18!$ possibilities. Same if Jana is at end. There are $18$ possibilities for the other places, $8\cdot7$ for her partners and $17!$ for the rest.

Answer:

$$\frac{2\cdot8\cdot18!+18\cdot56\cdot17!}{20!}$$

b. There are $\binom{20}4$ different commitee (with no restrictions). There are $\binom84$ commitees with boys and $\binom{12}4$ commitees with girls.

Answer:

$$1-\frac{\binom84+\binom{12}4}{\binom{20}4}$$