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I am a bit confused here. I am playing a game, and at the end of each match, 3 random players out of a total of 10 are awarded prizes What is the probability of me getting an item, considering any player may only get one item.

I'm sure it's not as easy as 3/10.

Thanks!

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    $\begingroup$ Yes, it is that easy. (If the three winners are selected at random, any three equally likely.) But you could do it the hard way. There are $\binom{10}{3}$ ways to select $3$ winners. There are $\binom{9}{2}$ ways to select $2$ people to join you on the podium. So the probability is $\binom{9}{2}$ divided by $\binom{10}{3}$. After simplifying, you will get $3/10$. $\endgroup$ May 8, 2014 at 17:06
  • $\begingroup$ If you want to get more proofy, there are $\binom{10}{3}=120$ ways to distribute the prizes. Assuming you get a prize, there are $\binom{9}{2}=36$ ways of distributing the others. $\frac{36}{120}=0.3$ $\endgroup$
    – RandomUser
    May 8, 2014 at 17:09
  • $\begingroup$ @Andre that's exactly what got me confused, thanks for clearing it out for me! Can you please give an actual answer so I can select it $\endgroup$ May 8, 2014 at 17:13

2 Answers 2

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Yes, it is that easy. (If the three winners are selected at random, any three equally likely.)

But you could do it the hard way. There are $\binom{10}{3}$ ways to select $3$ winners. There are $\binom{9}{2}$ ways to select $2$ people to join you on the podium. So the probability is $\binom{9}{2}$ divided by $\binom{10}{3}$. After simplifying, you will get $3/10$.

To justify the easy way, suppose there are three medals awarded, gold, silver, and plastic. Let $A$ be the event you win the gold medal, $B$ be the event you win the silver, and $C$ the event you win the plastic. Then the events $A$, $B$, and $C$ are pairwise disjoint. I think you will see that the probability you win gold is $\frac{1}{10}$. We get the same values for silver and for plastic. For the event $A\cup B\cup C$ that you win some medal, add.

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  • $\begingroup$ It all makes sense, thank you :) $\endgroup$ May 8, 2014 at 17:19
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Formally, there are ${9 \choose 2}$ ways for you to get an item and ${10 \choose 3}$ ways for the three items to be distributed, so $$ P(\text{you get an item}) = \frac{{9 \choose 2}}{{10 \choose 3}} = \frac{9! 7! 3!}{10! 7! 2!} = \frac{3}{10}. $$ However, there is an easier way to see this. Simply notice that everyone has an equal chance of being picked to win a prize, and the total expected number of prizes received by all payers is $3$. Therefore each player must have contributed an expected value of $\frac{3}{10}$ to the total expected number of prizes.

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  • $\begingroup$ That makes sense now, thank you! $\endgroup$ May 8, 2014 at 17:17

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