Lucky tickets

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Valentas

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There are 10 tickets from which 2 are winning. Randomly 5 tickets are taken. What is the probability that among taken tickets at least one is winning.

Answer is 7/9 but we haven't taken probabilities in advanced level and I believe it is from it. Could someone help me?

Thank you.
 
Valentas said:
There are 10 tickets from which 2 are winning. Randomly 5 tickets are taken. What is the probability that among taken tickets at least one is winning. Answer is 7/9.
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The combination \(\displaystyle \binom{8}{5}=\frac{8!}{5!\cdot 3!}\) is the number of ways that no winning ticket is among the five.

So the answer is \(\displaystyle 1-\dfrac{\binom{8}{5}}{\binom{10}{5}}\).

BTW that is the given answer.
 
JeffM said:
Questions that have "at least" or "at most" in them are often most easily solved indirectly.

The probability that no winner is picked OR AT LEAST one winner is picked is certainty, or 1. Make sense so far?

The probability that no winner is picked AND at least one winner is picked is nil, or 0. Make sense so far?

So P(no winner OR at least one winner) = P(no winner) + P(at least one winner) - P(no winner AND at least one winner).

So P(at least one winner) = 1 - P(no winner) + 0 = 1 - P(no winner).

How many ways can you pick 5 from 10 cards? (Do you know how to do this kind of computation?) The answer is 252.

How many ways can you pick 5 from the 8 non-winning cards. (Do you see why this is relevant?) The answer is 56.

So P(no winner) \(\displaystyle = \dfrac{56}{252} = \dfrac{7*8}{7*9*4} = \dfrac{2}{9}\)

P(at least one winner) \(\displaystyle = 1 - \dfrac{2}{9} = \dfrac{7}{9}\)

Did this help?
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Thank you, sir, this was extremely helpful. Your way of thinking impressed me. Thank you. :)
 
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