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In a student council, there are 8 first year students, 6 second year students, 5 third year students and 6 fourth grade students. 5 students will be randomly chosen as school representatives. All students have a equal chance of being a school representative.

A) What is the chance that 2 first year students and 1 student from each other grade become school representatives?

B) What is the chance that 3 second year students and 2 fourth year students become representatives?

The answer for A is $0.095$ and for B its $0.8056$ .

I thought to use combinations for choosing students and multiply the results, which in turn I would divide with the number of possible results as a whole, but it's giving me the wrong answers.

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  • $\begingroup$ The method you thought of using is good enough. And the same method gives the answers (approximately, of course) that you've mentioned. Maybe you are calculating something wrong? $\endgroup$ Aug 15 '20 at 8:07
  • $\begingroup$ Maybe its something about the sum of all possible results? ive run the same calculation several times for the multiplication part and its always the same, and im confident that its right for that part, but my sum of all results might be wrong, first i tried to use combination without repetition then i tried variation without repetition and so on and none seem to be close enough. $\endgroup$ Aug 15 '20 at 8:21
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In how many ways can you choose $2$ first year, $1$ second year, $1$ third year and $1$ fourth year students? You can make each of these choices in $\binom{8}{2}, \binom{6}{1}, \binom{5}{1}, \binom61$ ways, and since all these choices are independent, we can choose the representative group of $5$ people in this $(2,1,1,1)$ composition in $\binom82\cdot\binom61\cdot\binom51\cdot\binom61=5040$ ways (by the multiplication principle of counting) and the total number of ways we can choose $5$ students from $8+6+5+6=25$ students is $\binom{25}{5}$, so we the required probability you want is $$\dfrac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}=\dfrac{5040}{\binom{25}5}= 0.094861\cdots$$

Try the second part in the same way.

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  • $\begingroup$ I now know what i did wrong. insted of 25 i wrote 24 and never checked if tis wrong. Thank you. $\endgroup$ Aug 15 '20 at 8:36
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There are total $25$ students. So, no. of ways of choosing any $5$ out of them is $$n(S)={25\choose 5}$$ By the given conditions, $$n(A)={8\choose 2}{6\choose 1}{5\choose 1} {6\choose 1}$$ and $$n(B)={6\choose 3} {6\choose 2} $$

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