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Difficulty:
95%
(hard)
Question Stats:
28%
(02:50)
correct 72%
(02:53)
wrong
based on 221
sessions
History
Date
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Not Attempted Yet
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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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. \(\frac{1}{5}\)
Show more
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) =\(\frac{11}{125}*\frac{9}{10}=\frac{99}{1250}\)
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. \(\frac{1}{5}\)
Show more
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
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.
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?
The probability of Abe getting book published = 1/2*20% = 1/10 The probability of Abe not getting book published = 1 - 1/10 = 9/10
The probability of Beth getting book published = 1/5*20% = 1/25 The probability of Beth not getting book published = 1 - 1/25 = 24/25
The probability of Carl getting book published = 1/4*20% = 1/20 The probability of Carl not getting book published = 1 - 1/20 = 19/20
The probability that Beth or Carl getting book published = 1/25 + 1/20 - (1/25)(1/20) = 44/500 = 22/250
The probability that Beth or Carl, but not Abe, will get their book published = 22/250*9/10 = 99/1250
IMO C
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.