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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink]
Are you sure the answer is 0.6 and not 0.44 ?

P( at least 1 will hit ) = 1 - P ( none will hit )

And, P ( none will hit ) = P( A won't hit) * P (B won't hit)
= (1 - 0.3 ) * ( 1- 0.2)
= 0.7 * 0.8
= 0.56

Thus, P ( at least 1 will hit) = 1 - 0.56 = 0.44.
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink]
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OE

Quote:
This is a simultaneous/independent probability problem in that Hank and Babe's batting performances do not influence each others'. The probability that at least one player will hit a home run in today's game is (0.3)(0.2) = 0.6.
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink]
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Carcass wrote:
OE

Quote:
This is a simultaneous/independent probability problem in that Hank and Babe's batting performances do not influence each others'. The probability that at least one player will hit a home run in today's game is (0.3)(0.2) = 0.6.


But, 0.3*0.2 = 0.06
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink]
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krpajedrez wrote:
Carcass wrote:
OE

Quote:
This is a simultaneous/independent probability problem in that Hank and Babe's batting performances do not influence each others'. The probability that at least one player will hit a home run in today's game is (0.3)(0.2) = 0.6.


But, 0.3*0.2 = 0.06


True sir

You are perfectly correct. The book has a typo.

I edited the OA
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink]
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I was just pointing out your calculation error.

But your logic is still wrong.
The question asks for "at least one", and you are determining the probability of both simultaneously.
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink]
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krpajedrez wrote:
I was just pointing out your calculation error.

But your logic is still wrong.
The question asks for "at least one", and you are determining the probability of both simultaneously.

You're absolutely right. I have changed the official answer to 0.44
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink]
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We just subtract both not hitting a home run from 1 (total of all probabilities).
1-0.7*0.8=0.44.
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink]
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