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A group consisting of $3$ men and $6$ women attends a prizegiving ceremony. If $ 5$ prizes are awarded at random to members of the group, find the probability that exactly $3 $ of the prizes are awarded to women if
a) There is a restriction of at most one prize per person
b) There is no restriction on the number of prizes per person

I did part a) and got the same result as the solution but I failed at getting the same answer for part b). When I looked at the working outs of both parts, I noticed a significant difference in the ways two parts are solved.

This is the working out for part a) (which is also similar to my working out) a) $\frac{6C3\times 3C2}{9C5} = \frac{10}{21}\ $

And this is the working out of part b) b) $\ 5C3 \times (\frac{3}{9})^{2} \times (\frac{6}{9})^{3}\ = \frac{80}{243}\ $

I'm so confused why part b) is done in such a different way than part a) and as a student, how can I know when to consider the numerator and denominator separately like part a) and when to find the probability of each component and times all of them together like part b)? Also, can we solve part b) in a similar way like part a)? Does anyone have any tips on how to distinguish these sorts of methods?

Thank you very much for helping.

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    $\begingroup$ What is the answer for part b $\endgroup$ – Isaac YIU Math Studio Sep 21 '19 at 1:46
  • $\begingroup$ Hi, I just added the answer to part b) Thank you for taking the time looking at my question $\endgroup$ – BooScout Sep 21 '19 at 2:10
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b) There are $5$ independent events in the form of prizes that are awarded that can succeed each (i.e. the prize is awarded to a woman) with (the same) probability $\frac69$, or fail (i.e. the prize is not awarded to a woman).

So evidently we are dealing with binomial distribution here, equipped with parameters $n=5$ and $p=\frac39$.

Essential difference: in a) the events are not independent. If e.g. the first prize is handed over to Bob then for the further process Bob is put aside because he cannot be awarded with another prize.

Actually I would say that b) is more easy to solve than a) where we are dealing with hypergeometric distribution.

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Let's start.( Answer for b)

Total number of prizes $=5$.

The number of ways to select $3$ prizes from $5$ prizes $=5C3$ $=10$.

Given that $3$ prizes goes to women and $2$ prizes goes to men.

And given that,there are $6$ women and $3$ men. So the total number of persons $=9$.

Let's consider the scene, I am going to distribute $3$ prizes among $6$ womens (a person can get more than one prize).

So that, total number of possible ways to distribute $3$ prizes among $6$ womens $=6×6×6= 6^3$.

Therefore, the probability of the event that to distribute $3$ prizes among $6$ womens $={6^3}/{9^3}$.

Let's argue the same argument for the scene, that, distributing the $2$ prizes among $3$ mens.

So, Therefore, the probability of the event that to distribute $2$ prizes among $3$ mens $={3^2}/{9^2}$.

(NOTE: If you choose $3$ prizes for women then remaining $2$ prizes goes to mens. So only there are $5C3$ such possibilities.)

Thus,the required ways $=10× ({6^3}/{9^3})× ({3^2}/{9^2})$ $=80/243$.

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Part b) : Assuming the prizes are identical there are $${9\choose 1}+4*{9\choose 2}+6*{9\choose 3}+4*{9\choose 4}+{9\choose 5}=c={13\choose 5}$$ total ways of distributing them ; the nth term the count when there are n winners. Exactly $$({6\choose 3}+2*{6\choose 2}+6)*(2*{3\choose 2}+3)=504$$ of these ways are desirable ; the number of ways of distributing 3 to 6 then multiplied by the number of ways of distributing 2 to 3. So, the probability is $$504/c$$ $c$ is the number of ways 5 can be written as the sum of 9 non-negative integers with respect to order ; i.e. the number of non-negative integer solutions to $$\sum_{i=1}^9 x_i =5$$ i.e. the coefficient of $x^5$ in the polynomial $$(\sum_{i=0}^5 x^i)^9$$

In general the number of ways of distributing n identical objects to m people is $${{m+n-1}\choose {n-1}}$$

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If the prizes are not identical and each prize is distinct then the total # of outcomes in a) becomes $${9\choose 5}*5!$$ whereas in b) it becomes $9^5$ which counts the # of functions from a 5-set to a 9-set. The # of desirable outcomes in a) is then $${6\choose 3}*{5\choose 3}*3!*{3\choose 2}*2$$ yielding for a) the probability $\frac {{6\choose 3}*{3\choose 2}}{9\choose 5}=\frac {10}{21}$. The number of desirable outcomes in b) becomes $${5\choose 3}*6^3*3^2$$ because we choose what 3 prizes are awarded to the female winner(s) then multiply by the # of functions from a 3-set to a 6-set and finally multiply by the # of functions from a 2-set to a 3-set. This yields for b) the probability $$\frac {{5\choose 3}*6^3*3^2}{9^5}=\frac {80}{243}$$

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