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If the probabilities of Hank and Babe, two baseball players in the sam

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Carcass
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If the probabilities of Hank and Babe, two baseball players in the sam [#permalink] New post   19 Apr 2022, 01:00
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If the probability of Hank and Babe, two baseball players in the same MLB team, of hitting a home run in today's game is 0.3 and 0.2, respectively, what is the probability that at least one of them will hit a home run in today's game?

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0.44


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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink] New post   23 May 2022, 06:21
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Carcass wrote:
If the probability of Hank and Babe, two baseball players in the same MLB team, of hitting a home run in today's game is 0.3 and 0.2, respectively, what is the probability that at least one of them will hit a home run in today's game?

Show: ::
0.44


Given:
P(Hank hits home run) = 0.3, which means P(Hank does NOT hit home run) = 1 - 0.3 = 0.7
P(Babe hits home run) = 0.2, which means P(Babe does NOT hit home run) = 1 - 0.2 = 0.8

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(at least 1 home run) = 1 - P(not getting at least 1 home run)
So, we can write: P(getting at least 1 home run) = 1 - P(getting zero home runs)

P(getting zero home runs)
P(getting zero home runs) = (Hank does not get a home run AND Babe does not get a home run)
= (Hank does not get a home run) x P(Babe does not get a home run)
= 0.7 x 0.8
= 0.56

So, P(getting at least 1 home run) = 1 - 0.56 = 0.44
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink] New post   27 Apr 2022, 22:13
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Re: If the probabilities of Hank and Babe, two baseball players in the sam [#permalink] New post   23 May 2022, 00:35
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] New post   23 May 2022, 03:26
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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] New post   23 May 2022, 03:38
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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] New post   23 May 2022, 04:03
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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] New post   23 May 2022, 05:11
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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] New post   23 May 2022, 06:22
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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] New post   03 Jun 2022, 02:07
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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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