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Q. 6 - Q. 10 carry two marks each.
Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict is
- a)0.14
- b)0.22
- c)0.33
- d)0.40
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each...
X - A hits the convict
Y - B hits the convict
Given, P(X) = 3 * P(Y)
Z - Convict is injured
Z' - Convict is not injured
Given, P(Z') = 0.5
P(Z) = 1 - P(Z')
P(Z) = 1 - 0.5 = 0.5
Now,
P(Z) = P(X) * P(Y') + P(X') * P(Y) + P(X) * P(Y)
Let
P(Y) = t
P(X) = 3t
P(Y') = 1-t
P(X') = 1-3t
Substituting in above equation,
0.5 = (3t * (1-t)) + ((1-3t) * t) + (t * 3t)
=> 3t - 3t2 + t - 3t2 + 3t2 = 0.5
=> 3t2 - 4t + 0.5 = 0
=> 6t2 - 8t + 1 = 0
Solving, we get
t=1.193(eliminated as probability cannot be greater than 1) OR t=0.1396
Therefore. P(Y) = t = 0.1396
Answer A) 0.14
Most Upvoted Answer
Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each...
To solve this problem, we can use the concept of conditional probability. Let's break down the given information step by step.
Given:
- Probability of the convict not getting injured = 0.5
- Probability that A hits the convict is three times the probability that B hits the convict
Let's denote the probability that A hits the convict as P(A) and the probability that B hits the convict as P(B). According to the given information, we have:
P(A) = 3P(B)
We also know that the probability of the convict not getting injured is 0.5. This means that the probability of at least one of the policemen hitting the convict is 1 minus the probability that the convict does not get injured. Mathematically, we can express this as:
P(A or B) = 1 - 0.5
P(A or B) = 0.5
Now, we can use the concept of conditional probability to calculate P(B) using the formula:
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
Since both policemen fire at the same time, the probability that B hits the convict given that A has already hit the convict (B|A) is 0. So, we can simplify the formula to:
P(B) = P(B|not A) * P(not A)
We know that P(A) = 3P(B), so P(not A) = 1 - P(A). Substituting this in the formula, we get:
P(B) = P(B|not A) * (1 - P(A))
Now, we need to find the value of P(B|not A). This can be calculated using the fact that P(A) + P(not A) = 1. Rearranging this equation, we get:
P(not A) = 1 - P(A)
Substituting this in the formula for P(B), we get:
P(B) = P(B|not A) * P(A)
We know that P(A) = 3P(B), so the equation becomes:
P(B) = P(B|not A) * 3P(B)
Simplifying further, we get:
1 = P(B|not A) * 3
Therefore, P(B|not A) = 1/3.
Now, substituting this value in the formula for P(B), we get:
P(B) = (1/3) * P(A)
Since P(A) = 3P(B), the equation becomes:
P(B) = (1/3) * 3P(B)
Simplifying further, we get:
P(B) = P(B)
This implies that the probability of B hitting the convict is equal to the probability of B hitting the convict given that A has not hit the convict.
Since P(B) + P(not B) = 1, we can write:
P(not B) = 1 - P(B)
Substituting the value of P(B) we obtained earlier, we get:
P(not B) = 1 - (1/3) * P(A)
We know that P(A or B) = 0.5. Substituting the values of P(A) and P(B), we get:
0.5 = P(A) + P(B) - P(A and B)
0.
Given:
- Probability of the convict not getting injured = 0.5
- Probability that A hits the convict is three times the probability that B hits the convict
Let's denote the probability that A hits the convict as P(A) and the probability that B hits the convict as P(B). According to the given information, we have:
P(A) = 3P(B)
We also know that the probability of the convict not getting injured is 0.5. This means that the probability of at least one of the policemen hitting the convict is 1 minus the probability that the convict does not get injured. Mathematically, we can express this as:
P(A or B) = 1 - 0.5
P(A or B) = 0.5
Now, we can use the concept of conditional probability to calculate P(B) using the formula:
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
Since both policemen fire at the same time, the probability that B hits the convict given that A has already hit the convict (B|A) is 0. So, we can simplify the formula to:
P(B) = P(B|not A) * P(not A)
We know that P(A) = 3P(B), so P(not A) = 1 - P(A). Substituting this in the formula, we get:
P(B) = P(B|not A) * (1 - P(A))
Now, we need to find the value of P(B|not A). This can be calculated using the fact that P(A) + P(not A) = 1. Rearranging this equation, we get:
P(not A) = 1 - P(A)
Substituting this in the formula for P(B), we get:
P(B) = P(B|not A) * P(A)
We know that P(A) = 3P(B), so the equation becomes:
P(B) = P(B|not A) * 3P(B)
Simplifying further, we get:
1 = P(B|not A) * 3
Therefore, P(B|not A) = 1/3.
Now, substituting this value in the formula for P(B), we get:
P(B) = (1/3) * P(A)
Since P(A) = 3P(B), the equation becomes:
P(B) = (1/3) * 3P(B)
Simplifying further, we get:
P(B) = P(B)
This implies that the probability of B hitting the convict is equal to the probability of B hitting the convict given that A has not hit the convict.
Since P(B) + P(not B) = 1, we can write:
P(not B) = 1 - P(B)
Substituting the value of P(B) we obtained earlier, we get:
P(not B) = 1 - (1/3) * P(A)
We know that P(A or B) = 0.5. Substituting the values of P(A) and P(B), we get:
0.5 = P(A) + P(B) - P(A and B)
0.
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Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each...
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Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer?
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Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer?.
Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer?.
Solutions for Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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ample number of questions to practice Q. 6- Q. 10carry two marks each.Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict isa)0.14b)0.22c)0.33d)0.40Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice GATE tests.
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