[Math] Probability question – arranging 20 pupils in a row – 8 boys and 12 girls

combinatoricspermutationsprobabilityprobability theory

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!

Best Answer

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:

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

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