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Roy is making plans for the evening. He can watch a movie,

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Roy is making plans for the evening. He can watch a movie, [#permalink] New post 30 Sep 2003, 13:48
Roy is making plans for the evening. He can watch a movie, go to a bar, or study for the MATH250 test. He is twice as likely to watch a movie than to go to a bar, and equally likely to watch a movie as he is to study for the test. What is the probability that Roy will study for the test?
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Re: Probability # 2 [#permalink] New post 01 Oct 2003, 07:08
Prob. that Roy watches a movie = P(M)
Prob. that Roy goes to a bar = P(B)
Prob. that Roy studies for the test = P(S)

P(M) + P(B) + P(S) = 1 ........ (1)

Given, P(M) = 2 * P(B)
P(M) = P(S)

So, (1) can be rewritten as P(S) + (P(S) / 2) + P(S) = 1
i.e. P(S) * 5/2 = 1 => P(S) = 2/5
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Re: Probability # 2 [#permalink] New post 04 Oct 2003, 05:22
amarsesh wrote:
Prob. that Roy watches a movie = P(M)
Prob. that Roy goes to a bar = P(B)
Prob. that Roy studies for the test = P(S)

P(M) + P(B) + P(S) = 1 ........ (1)

Given, P(M) = 2 * P(B)
P(M) = P(S)

So, (1) can be rewritten as P(S) + (P(S) / 2) + P(S) = 1
i.e. P(S) * 5/2 = 1 => P(S) = 2/5


Is there any other approach to such problems? just in case. i also got 2/5
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 [#permalink] New post 04 Oct 2003, 08:09
My approach is employing a basic formula P=F/T.
Let: P(bar) = p
Then: P(movie) = 2p; P(test) = 2p

The total set of variants = 5p
The favorable set = 2p

P=2p/5p=2/5
  [#permalink] 04 Oct 2003, 08:09
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