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Frederico is going to try walk across the United States of America. He
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18 Oct 2021, 05:30
18 Oct 2021, 05:30
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Frederico is going to try walk across the United States of America. He has decided that he will stop walking at the end of the first day that his knee starts to cause him pain. If the probability of a knee injury that would cause Frederico pain each day is .4, what is the probability of him ending his walk at the end of the third day?
A. 0.144
B. 0.216
C. 0.64
D. 0.96
E. 0.6
A. 0.144
B. 0.216
C. 0.64
D. 0.96
E. 0.6
Re: Frederico is going to try walk across the United States of America. He
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18 Oct 2021, 06:20
18 Oct 2021, 06:20
1
The probability of not pain in the first day is 1 - 0.4 = 0.6, similar for the second day.
The probability he feels pain on the third day is (not pain on 1st day) x (not pain on 2nd day) x (pain on 3rd day) = 0.6 x 0.6 x 0.4 = 0.144
Choice A
The probability he feels pain on the third day is (not pain on 1st day) x (not pain on 2nd day) x (pain on 3rd day) = 0.6 x 0.6 x 0.4 = 0.144
Choice A
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Re: Frederico is going to try walk across the United States of America. He
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18 Oct 2021, 06:46
18 Oct 2021, 06:46
1
Carcass wrote:
Frederico is going to try walk across the United States of America. He has decided that he will stop walking at the end of the first day that his knee starts to cause him pain. If the probability of a knee injury that would cause Frederico pain each day is .4, what is the probability of him ending his walk at the end of the third day?
A. 0.144
B. 0.216
C. 0.64
D. 0.96
E. 0.6
A. 0.144
B. 0.216
C. 0.64
D. 0.96
E. 0.6
If P(knee pain) = 0.4, then P(NO knee pain) = 0.6
P(Fred ends walk at end of Day 3) = P(no knee pain on Day 1 AND no knee pain on Day 2 AND knee pain on Day 3)
= P(NO knee pain on Day 1) x P(NO knee pain on Day 2) x P(knee pain on Day 3)
= 0.6 x 0.6 x 0.4
= 0.144
Answer: A
