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You have to choose a committee of $5$ people. You have $13$ grade nines, $13$ grade $10s$, $13$ grade elevens and $13$ grade $12s$.

$a)$ What is the probability that the committee chosen will be all of the same grade?

$b)$ What is the probability that the committee will be from two grades. $3$ from one grade, and $2$ from another grade?

Attempt:

$a)$

= $\dfrac{\large ^{13}C_5}{\large^{52}C_5} * 4$ (four different grades)

$b)$

$= \dfrac{^{13}C_2 *^{13}C_3}{\large ^{52}C_5} * 3!2!$ ($12$ different combinations)

would my answers be wrong?

Apurv
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Jessica
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2 Answers2

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To get the number of ways to get five students all of the same grade, first pick the grade, which is ${4 \choose 1}$, and then pick the students from that grade, which is ${13 \choose 5}$. Then the probability of this happening is:

$$P_a = \frac{{4 \choose 1}{13 \choose 5}}{{52 \choose 5}}.$$

So I agree with your first one.

To get the second one, first pick the grade that has the pair, which is ${4 \choose 1}$, and then pick the grade that has the group of three: ${3 \choose 1}$. Then pick the pair: ${13 \choose 2}$. Then pick the trio: ${13 \choose 3}$.

$$P_b = \frac{{4 \choose 1}{3 \choose 1}{13 \choose 2}{13 \choose 3}}{{52 \choose 5}}.$$

So I agree with your second one as well. Rock on.

John
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  • Thank you!!! But does the permutation in part b make sense? I meant the 3!2! as the different combos of the pair and the group of three – Jessica Jan 29 '14 at 05:22
  • I don't see where you're getting that from. Both $3!2!$ and ${4 \choose 1}{3 \choose 1}$ work out to $12$, but what was your reasoning for doing ($3!2!$)? I'm not saying (yet) that you're wrong, just that I don't understand your thought process. – John Jan 29 '14 at 05:43
  • Because imagine a combination of three of the same grade, and 2 of the same grade, for example, G = grade where the three are coming from, T = grade where the other two are coming from, therefore its a combination if GGGTT, so 3!2! to get the combos – Jessica Jan 31 '14 at 02:27
  • Ahhh, OK. This is a completely different issue. What you calculated was the number of ways to line them up, assuming that you have the three $G$'s on the left and the two $T$'s on the right. There are six ways to arrange the $G$'s: $123, 132, 213, 231, 312, 321$. There are two ways to arrange the $T$'s: $12, 21$. But that's not at all what was needed here. It was just a lucky coincidence that the numbers worked out the same. :) – John Jan 31 '14 at 02:48
  • haha i see.. so the 3!2! would be wrong? – Jessica Jan 31 '14 at 03:14
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For a, you are doing fine. It would be easier to check if you added some words about what your thinking is. For b, the numeric answer is correct, but I don't understand the logic of the $3!2!=12$ I would have said there are four choices for the grade that supplied two, then (since it has to be different) three choices for the grade that supplied three. That also gives a product of $12$.

Ross Millikan
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  • 3!2! represent the combinations of the pair and the group of three – Jessica Jan 29 '14 at 05:10
  • As in the number of different combinations of two of the same grade, and three (eg. 2 grade nine & 3 grade ten[1], 2 grade nine & 3 grade elevens[2], 2 grade nine & 3 grade twelve[3]....and so on) – Jessica Jan 29 '14 at 05:13
  • No, you got the combinations of two of the first grade with your $^{13}C_2$ and the combinations of three of the second grade with ${13}C_3$. Now you are picking the grades. If there were five grades with $12$ students in each, the solution would be$ ^{13}C_2^{13}C_3\cdot 5 \cdot 4$, while you would still be multiplying by $3!2!$ – Ross Millikan Jan 29 '14 at 13:57