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23 Aug 2022, 01:18
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Difficulty:
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A biased die has a probability of 1/4 of showing a 5, while the probability of any of 1,2,3,4 or 6 turning up is the same. If three such dice are rolled, what is the probability of getting a sum of at least 14 without getting a 6 on any die?
A. 3/40
B. 1/30
C. 13/120
D. 7/160
E. 1/20
(adapted from gmatfree)
A. 3/40
B. 1/30
C. 13/120
D. 7/160
E. 1/20
(adapted from gmatfree)
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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23 Aug 2022, 05:35
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Since 5 has a 1/4 chance the remaining numbers each have:
3/(4*5) = 3/20
The only way to roll as stipulated is:
555 and 545
Probability of three 5's is:
(1/4)^3 = 1/64 = 5/320
There are 3 ways to roll 545:
3*(1/4)^2*3/20 = 9/320
Total probability:
5/320 + 9/320 = 14/320 =
7/160
Posted from my mobile device
3/(4*5) = 3/20
The only way to roll as stipulated is:
555 and 545
Probability of three 5's is:
(1/4)^3 = 1/64 = 5/320
There are 3 ways to roll 545:
3*(1/4)^2*3/20 = 9/320
Total probability:
5/320 + 9/320 = 14/320 =
7/160
Posted from my mobile device
23 Aug 2022, 23:52
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pintukr
A biased die has a probability of 1/4 of showing a 5, while the probability of any of 1,2,3,4 or 6 turning up is the same. If three such dice are rolled, what is the probability of getting a sum of at least 14 without getting a 6 on any die?
A. 3/40
B. 1/30
C. 13/120
D. 7/160
E. 1/20
(adapted from gmatfree)
A. 3/40
B. 1/30
C. 13/120
D. 7/160
E. 1/20
(adapted from gmatfree)
Solution:
We are given \(P(5)=\frac{1}{4}\)
Thus, the probability of the rest of them \(P(1)+P(2)+P(3)+P(4)+P(6)=1-\frac{1}{4}=\frac{3}{4}\)
Or \(P(1)=P(2)=P(3)=P(4)=P(6)=3/4/5=\frac{3}{20}\)
We have three of these dice and want the sum of at least 14 without getting 6. We will have the following scenarios:
Attachment:
probability.png
We cannot have a sum 16, 17 or 18 without getting a 6
Thus, total probability \(= \frac{3}{320}+\frac{3}{320}+\frac{3}{320}+\frac{1}{64}\)
\(⇒\frac{3+3+3+5}{320}\)
\(⇒\frac{14}{320}\)
\(⇒\frac{7}{160}\)
Hence the right answer is Option D
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.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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.
Thank you for understanding, and happy exploring!
