QPROBLEM 7 : 2.0 POINTS
A school is running a raffle. There are 100 tickets, of which 3 are winners. You can assume that tickets are sold by drawing at random without replacement from the available tickets. Teacher X buys 10 raffle tickets, and so does Teacher Y. Find the chance that one of those two teachers gets all three winning tickets.

- anonymous

- katieb

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- anonymous

okay that was wrong, i did not read carefully

- anonymous

lets compute the probability that teacher X gets all three winners.
there are \(\binom{100}{10}\) ways to pick 10 out of 100. if it includes the three winning tickets then there are \(\binom{3}{3}=1\) way for them to be included and \(\binom{97}{7}\) ways to choose the other 7, so the probability that teacher X gets all three is
\[\frac{\binom{97}{3}}{\binom{100}{10}}\]

- anonymous

damn typo
last line should be
\[\frac{\binom{97}{7}}{\binom{100}{10}}\]

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## More answers

- anonymous

this is the same as the probability that teacher Y gets all the winners, and since these events are clearly disjoint (they can't both have all the winners) you can get the probability that one or the other gets all the winners doubling the above answer

- kropot72

The probability that the 20 tickets bought between the 2 teachers contain the 3 winners is
\[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}\]
Given that the 20 tickets bought between the 2 teachers include the 3 winners, the probability that one of the teachers has the 3 winners is
\[\frac{\left(\begin{matrix}17 \\ 7\end{matrix}\right)}{\left(\begin{matrix}20 \\ 10\end{matrix}\right)}\]
The required probability is
\[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}\times \frac{\left(\begin{matrix}17 \\ 7\end{matrix}\right)}{\left(\begin{matrix}20 \\ 10\end{matrix}\right)}\]

- anonymous

What is the easiest way to calculate the probability for this problem in decimals? Is there some online calculator that could simplify and accelerate the process of caclulation?
Thanx.

- kropot72

A calculator with the factorial function will be helpful. For example
\[\left(\begin{matrix}97 \\ 17\end{matrix}\right)=\frac{97!}{17!80!}=\frac{97\times 96\times 95\times 94\times ....\times 89\times 88\times 87\times 86\times 85\times 84\times 83\times 82\times 81}{17!}\]

- anonymous

I have found online caclulator with factorial function, and, using the above given formula for the probability that 20 tickets bought between the 2 teachers contain the 3 winners I got the number 0,7045. Could someone just please double-check if I got the correct answer?
Thx

- kropot72

Your result is a long way from being correct.

- anonymous

Well, what is then the correct answer? And could you please explain step by step how you got it? Take some other numbers if you want, I just want to understand the principle. Many thanx.

- anonymous

\[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}\times \frac{\left(\begin{matrix}17 \\ 7\end{matrix}\right)}{\left(\begin{matrix}20 \\ 10\end{matrix}\right)}=0.000742115\]
m using excel for calculation...
is that correct ans?????

- anonymous

@kropot72 @njux

- anonymous

@abhi_abhi I have no idea, I am still trying to figure out the easiest way to do all those factorial calculations you have already done.
I hope someone will be able to confirm whether you have the right answer or not.
Best,

- anonymous

so what is the answer

- anonymous

i checked it its not correct its wrong guys

- anonymous

kropot72 whats wrong with you

- kropot72

The calculation by @abhi_abhi gives the same result as I get (0.000742). Have you tried the result of the method posted by @satellite73 as follows:
\[2\times \frac{\left(\begin{matrix}97 \\ 7\end{matrix}\right)}{\left(\begin{matrix}100 \\ 10\end{matrix}\right)}\]

- anonymous

well the my answer is 0.0015

- anonymous

and its right

- anonymous

NOP ITS WRONG

- kropot72

@JULIAKAPRI Have you posted your calculation. If not would you care to do so?

- kropot72

My previous attempt at a solution is not correct. I believe the following method gives a valid solution.
Assuming that the teachers' tickets are drawn first, the probability that the 20 tickets bought between the 2 teachers contain the 3 winners is
\[\frac{\left(\begin{matrix}97 \\ 17\end{matrix}\right)}{\left(\begin{matrix}100 \\ 20\end{matrix}\right)}=0.00705\]
Given that the 20 tickets bought between the 2 teachers include the 3 winners, the probability that one of the teachers has the 3 winners is can be found from the binomial distribution. The probability of a random ticket being a winner is 3/20. the probability of exactly 3 tickets out of a sample of 10 tickets is given by
\[P(3\ winners\ out\ of\ 10\ tickets)=\left(\begin{matrix}10 \\ 3\end{matrix}\right)(0.15)^{3}(0.85)^{7}=0.129834\]
The probability that one of the two teachers gets all three winning tickets is therefore
0.00705 * 0.129834 = 0.000915

- anonymous

N=100, G=3, n=10, g=3
probability teacher X gets the 3 winning tickets is: ((G,g)(N-G,n-g)/(N,n))
probability teacher Y gets the 3 winning tickets is: ((G,g)(N-G,n-g)/(N,n))
probability either teacher X or teacher Y gets the 3 winning tickets is: ((G,g)(N-G,n-g)/(N,n))+((G,g)(N-G,n-g)/(N,n))

- anonymous

@hlpwntd, does it mean that @satellite73's answer (see above) is correct?

- anonymous

Looks pretty good to me

- anonymous

@hlpwntd, the result I got after doing the calculation is 0,00148423. How does it look to you? :)
Thx.

- anonymous

0.0015

- anonymous

Looks pretty good to me, rounded or not, depending on if you had to round or not

- anonymous

@hlpwntd and @abhi_abhi, thanks!

- anonymous

0.0015 is the correct answer

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