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08 Feb 2024, 13:55
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D
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Difficulty:
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Question Stats:
45% (02:29) correct
55%
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wrong
based on 118
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Paul participates in a shooting competition. Paul occasionally hits the target and occasionally misses the target. The probability of Paul hitting the target in any of the shots is x. What is the probability that in 3 shots, he hits the target at least two times?
(1) In the 3 shots he attempted, the probability that Paul misses the target at most two times is 0.992
(2) In the 3 shots he attempted, the probability that Paul misses the target at least two times is 0.104
(1) In the 3 shots he attempted, the probability that Paul misses the target at most two times is 0.992
(2) In the 3 shots he attempted, the probability that Paul misses the target at least two times is 0.104
08 Feb 2024, 14:12
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Paul participates in a shooting competition. Paul occasionally hits the target and occasionally misses the target. The probability of Paul hitting the target in any of the shots is x. What is the probability that in 3 shots, he hits the target at least two times?
Firstly, note that there are only the following four cases possible when shooting three times:
The question asks to find the combined probability of the first two cases: P(HHM) + P(HHH).
(1) In the 3 shots he attempted, the probability that Paul misses the target at most two times is 0.992.
The above implies that the combined probability of the first three cases, HHH, HHM, and HMH, is 0.992. Therefore, the probability of the fourth case, MMM, is 1 - 0.992 = 0.008. Hence, (1 - x)^3 = 0.008, which yields 1 - x = 0.2, and finally, x = 0.8. Knowing the probability, we can get P(HHM) + P(HHH). Sufficient.
(2) In the 3 shots he attempted, the probability that Paul misses the target at least two times is 0.104.
The above implies that the combined probability of the last two cases is 0.104. Therefore, the combined probability of the first two cases is 1 - 0.104 = 0.896. Sufficient.
Answer: D.
Firstly, note that there are only the following four cases possible when shooting three times:
- HHH
HHM
HMM
MMM
The question asks to find the combined probability of the first two cases: P(HHM) + P(HHH).
(1) In the 3 shots he attempted, the probability that Paul misses the target at most two times is 0.992.
The above implies that the combined probability of the first three cases, HHH, HHM, and HMH, is 0.992. Therefore, the probability of the fourth case, MMM, is 1 - 0.992 = 0.008. Hence, (1 - x)^3 = 0.008, which yields 1 - x = 0.2, and finally, x = 0.8. Knowing the probability, we can get P(HHM) + P(HHH). Sufficient.
(2) In the 3 shots he attempted, the probability that Paul misses the target at least two times is 0.104.
The above implies that the combined probability of the last two cases is 0.104. Therefore, the combined probability of the first two cases is 1 - 0.104 = 0.896. Sufficient.
Answer: D.
08 Feb 2024, 14:36
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Option D seems correct
Excellent question.
let's say=>
h = hit
m = miss
P(h) = x
P(m) = 1-x
Paul gives 3 shots. In these 3 shoits there can be 4 possiblities.
Statement 1:
probability that Paul misses the target at most two times is 0.992
that means |m| = number of times a miss happens is at most 2 (0 or 1 or 2)
So, P(\(|m|\leq{2}\))=0.992
P(mmm)=0.008
So, P(m)=\(\sqrt[3]{0.008}\) = 0.2
=> P(h) = x = 1-0.2 = 0.8
So, we need P(hhh or hhm) = 1-(P(mmm)+P(mmh))
Now P(mmh) = 3*0.2*0.2*0.8 = 0.096 [ because there are 3 cases for mmh likely mmh, mhm, hmm]
P(\(h\geq{2}\)) = 1-(0.008+0.096) = 1-0.104 = 0.896
Hence, Sufficient
Statement 2:
In the 3 shots he attempted, the probability that Paul misses the target at least two times is 0.104
This is nothing but what we calculated in the previous statement.
Hence, Sufficient
So, either of the statements is individually sufficient.
Hence, option D.
Excellent question.
let's say=>
h = hit
m = miss
P(h) = x
P(m) = 1-x
Paul gives 3 shots. In these 3 shoits there can be 4 possiblities.
| Hits | Miss | Cases |
|---|---|---|
| 3 | 0 | hhh |
| 2 | 1 | hhm, hmh, mhh |
| 1 | 2 | hmm, mhm, mmh |
| 0 | 3 | mmm |
Statement 1:
probability that Paul misses the target at most two times is 0.992
that means |m| = number of times a miss happens is at most 2 (0 or 1 or 2)
So, P(\(|m|\leq{2}\))=0.992
P(mmm)=0.008
So, P(m)=\(\sqrt[3]{0.008}\) = 0.2
=> P(h) = x = 1-0.2 = 0.8
So, we need P(hhh or hhm) = 1-(P(mmm)+P(mmh))
Now P(mmh) = 3*0.2*0.2*0.8 = 0.096 [ because there are 3 cases for mmh likely mmh, mhm, hmm]
P(\(h\geq{2}\)) = 1-(0.008+0.096) = 1-0.104 = 0.896
Hence, Sufficient
Statement 2:
In the 3 shots he attempted, the probability that Paul misses the target at least two times is 0.104
This is nothing but what we calculated in the previous statement.
Hence, Sufficient
So, either of the statements is individually sufficient.
Hence, option D.
