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# In a raffle in which 200 tickets are sold, there are 3

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Joined: 18 Aug 2005
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In a raffle in which 200 tickets are sold, there are 3 [#permalink]

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06 Nov 2005, 18:15
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In a raffle in which 200 tickets are sold, there are 3 prizes - first prize of \$100, second prize of \$50, third prize of \$30. A girl has one ticket in the raffle. What is the probability that she wins

(a) a prize,
(b) at least \$50.

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Joined: 23 Oct 2005
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06 Nov 2005, 18:35
sandman wrote:
In a raffle in which 200 tickets are sold, there are 3 prizes - first prize of \$100, second prize of \$50, third prize of \$30. A girl has one ticket in the raffle. What is the probability that she wins

(a) a prize,
(b) at least \$50.

I really suck at Probaility and counting. But here's my try.

1) Atleast a prize:

P(First) = 1/200 (she has one ticket, hence the probability of her winning the prize is 1 in 200)
P (Second) = 199/200 * 1/199 (Probability of not winning first * Probability that she wins 2nd)
P(third) = 199 / 200 * 198/199 * 1/198 (Prob of not winnning first and second * prob of winning third)

P (at least one) = P(First) + P(second) + P(Third)

2) Winning 3rd - P(Third) from above.

Am I right?

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06 Nov 2005, 19:07
Part a.
P(girl wins atleast 1 prize) = 1 â€“ P(girl wins no prizes)

P(no prizes) = P(no first prize) x P(no second prize) x P(no third prize)
= 199/200 x 198/199 x 197/198
= 197/200

So, P(girl wins atleast 1 prize) = 1 â€“ 197/200
= 3/200

Part b.
P(atleast \$50) = P(wins first prize) OR P(wins second prize)

P(wins first prize) = 1/200

P(winning second prize) = P(not winning first) AND P(she wins 2nd)
= 199/200 x 1/199
= 1/200

So, P(atleast \$50) = 1/200 + 1/200
= 1/100

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06 Nov 2005, 19:07
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# In a raffle in which 200 tickets are sold, there are 3

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