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Abe, Beth and Carl each start to write a science fiction book.

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Abe, Beth and Carl each start to write a science fiction book. [#permalink] Abe, Beth and Carl each start to write a science fiction book.   11 Sep 2021, 11:17
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Abe, Beth and Carl each start to write a science fiction book. Their individual probabilities for completing their respective books are \(\frac{1}{2}\), \(\frac{1}{5}\) and \(\frac{1}{4}\). A completed science fiction book has a 20% chance of getting published. What is the probability that Beth or Carl, but not Abe, will get their book published?

A. \(\frac{11}{250}\)

B. \(\frac{9}{200}\)

C. \(\frac{99}{1250}\)

D. \(\frac{81}{1000}\)

E. \(\frac{1}{5}\)
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Re: Abe, Beth and Carl each start to write a science fiction book. [#permalink] Abe, Beth and Carl each start to write a science fiction book.   12 Sep 2021, 05:36
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Kushchokhani
Abe, Beth and Carl each start to write a science fiction book. Their individual probabilities for completing their respective books are \(\frac{1}{2}\), \(\frac{1}{5}\) and \(\frac{1}{4}\). A completed science fiction book has a 20% chance of getting published. What is the probability that Beth or Carl, but not Abe, will get their book published?

A. \(\frac{11}{250}\)

B. \(\frac{9}{200}\)

C. \(\frac{99}{1250}\)

D. \(\frac{81}{1000}\)

E. \(\frac{1}{5}\)


Let us find the probability of each of them.

A: P to finish book \(\frac{1}{2}\), and 20% chance of this for publication => \(\frac{20}{100}*\frac{1}{2}=\frac{1}{10}\)
Thus P of not publishing the book = \(1-\frac{1}{10}=\frac{9}{10}\)

B: P to finish book \(\frac{1}{5}\), and 20% chance of this for publication => \(\frac{20}{100}*\frac{1}{5}=\frac{1}{25}\)

C: P to finish book \(\frac{1}{4}\), and 20% chance of this for publication => \(\frac{20}{100}*\frac{1}{4}=\frac{1}{20}\)

P of (B or C) = \(\frac{1}{20}+\frac{1}{25}-\frac{1}{20}*\frac{1}{25}=\frac{11}{125}\)

Thus P(B or C but not A) =\(\frac{11}{125}*\frac{9}{10}=\frac{99}{1250}\)


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Re: Abe, Beth and Carl each start to write a science fiction book. [#permalink] Abe, Beth and Carl each start to write a science fiction book.   12 Sep 2021, 07:08
Kushchokhani
Abe, Beth and Carl each start to write a science fiction book. Their individual probabilities for completing their respective books are \(\frac{1}{2}\), \(\frac{1}{5}\) and \(\frac{1}{4}\). A completed science fiction book has a 20% chance of getting published. What is the probability that Beth or Carl, but not Abe, will get their book published?

A. \(\frac{11}{250}\)

B. \(\frac{9}{200}\)

C. \(\frac{99}{1250}\)

D. \(\frac{81}{1000}\)

E. \(\frac{1}{5}\)

THe probability for B or C to complete the book but not both = 1/20 + 1/25 - 1/20*1/25 = 11/125

THe probability for A not to complete = 1-1/10 = 9/10

Therefore the required probability = 11 *9 /1250 = 99/1250

Therefore IMO C
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Re: Abe, Beth and Carl each start to write a science fiction book. [#permalink] Abe, Beth and Carl each start to write a science fiction book.   08 Feb 2022, 07:16
What if the question asked, "What is the probability that Beth or Carl will get their book published?"

Would we still subtract the "Both" part ? I am confused about why the "Both" part is even needed?

chetan2u

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Re: Abe, Beth and Carl each start to write a science fiction book. [#permalink] Abe, Beth and Carl each start to write a science fiction book.   08 Feb 2022, 23:37
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Mugdho
What if the question asked, "What is the probability that Beth or Carl will get their book published?"

Would we still subtract the "Both" part ? I am confused about why the "Both" part is even needed?

chetan2u

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Hi

You subtract BOTH when there is possibility of both happening together and is included in each case.

For example
When you choose P of Beth, it will contain when B publishes and C does not, and also when both B and C publish their books.
When you choose P of C, it will contain when C publishes and B does not, and also when both B and C publish their books.
So, both cases above contain 'both B and C', so it has to be subtracted once from total.
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Re: Abe, Beth and Carl each start to write a science fiction book. [#permalink] Abe, Beth and Carl each start to write a science fiction book.   14 Aug 2024, 11:52
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chetan2u

I have a small doubt with this question, say shouldn't the probability for not publishing be 1/2 again? As, that will also signify 1. did not even finish the book to get published 2. an incomplete book cannot be published.

I hope my doubt is clear, please do let me know. Thank you.
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Re: Abe, Beth and Carl each start to write a science fiction book. [#permalink] Abe, Beth and Carl each start to write a science fiction book.   04 Sep 2024, 12:48
Bunuel or chetan2u ,can one of you help me verify what I might be missing in my approach -

B(publishes) * C (does not publish) * A (does not publish) + B (does not publish) * C (publishes) * A (does not publish)

So :
(1/5 * 1/5) * (3/4 (does not complete) + 1/4 * 4/5) * (1/2 + 1/2 * 4/5) (case 1)
+
(4/5 + 1/5 * 4/5) * (1/4 * 1/5) * (1/2 + 1/2 * 4/5)

= 387/5000

What scenario am i overlooking / considering double to get a different answer than the OA ?
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Re: Abe, Beth and Carl each start to write a science fiction book. [#permalink] Abe, Beth and Carl each start to write a science fiction book.   05 Sep 2024, 00:11
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shaurya_gmat
Bunuel or chetan2u ,can one of you help me verify what I might be missing in my approach -

B(publishes) * C (does not publish) * A (does not publish) + B (does not publish) * C (publishes) * A (does not publish)

So :
(1/5 * 1/5) * (3/4 (does not complete) + 1/4 * 4/5) * (1/2 + 1/2 * 4/5) (case 1)
+
(4/5 + 1/5 * 4/5) * (1/4 * 1/5) * (1/2 + 1/2 * 4/5)

= 387/5000

What scenario am i overlooking / considering double to get a different answer than the OA ?
­B(publishes) * C (publishes) * A (does not publish)
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Re: Abe, Beth and Carl each start to write a science fiction book. [#permalink]
05 Sep 2024, 00:11
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