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The probability of a shooter hitting a target is 0.75. How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?
- a)4
- b)2
- c)3
- d)1
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The probability of a shooter hitting a target is 0.75. How many minim...
Given
Probability of success, p = 0.75
Concept used:
The binomial distribution formula is
P(X = x) = ncx qn – x px
Where,
P(X) = Probability of a success on an individual trial
p = Probability of successes
x = Number of success
q = Probability of failure or q = 1 – p
n = Number of trials
ncx = n!/(n – x)!x!
Calculation:
Here, p = 0.75 = (75/100) = ¾
So, probability of missed target = q = 1 – ¾
⇒ ¼
P(X = r) = ncx qn – x px, x = 0, 1, 2, ...., n
⇒ ncx(1/4)n – x (3/4)x = ncx(3x/4n)
Now, P(hitting the target at least once) > (1 – 0.01) = 0.99
⇒ P(x ≥ 1) > 0.99
So,1 – P (x = 0) > 0.99
⇒1 – nc0(1/4n) > 0.99
⇒ nc0(1/4n) < (1="" –="" />
⇒ nc0(1/4n) < />
⇒ (1/4)n < />
⇒ 4n > 1/0.01
⇒ 4n > 1/(1/100)
⇒ 4n > 100
The minimum value of n to satisfy the inequality (1) is 4.
∴ The shooter must fire 4 times
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Community Answer
The probability of a shooter hitting a target is 0.75. How many minim...
Required Probability: 0.99
Probability of hitting the target in one shot: 0.75
Let's assume the shooter fires n shots. The probability of not hitting the target in one shot is 1 - 0.75 = 0.25.
To find the probability of not hitting the target in n shots, we multiply the probability of not hitting the target in one shot by itself n times (since the shots are independent events).
P(not hitting the target in n shots) = (0.25)^n
The probability of hitting the target at least once in n shots is the complement of not hitting the target in n shots, so:
P(hitting the target at least once in n shots) = 1 - P(not hitting the target in n shots) = 1 - (0.25)^n
We want this probability to be more than 0.99, so:
1 - (0.25)^n > 0.99
Solving this inequality for n:
0.01 > (0.25)^n
n * log(0.25) > log(0.01)
n * log(0.25) > log(0.01)
n > log(0.01) / log(0.25)
Using a calculator, we find:
n > -2 / -0.6021
n > 3.32
Since n must be a whole number, the minimum number of times the shooter must fire is 4.
Therefore, the correct answer is option A) 4.
Probability of hitting the target in one shot: 0.75
Let's assume the shooter fires n shots. The probability of not hitting the target in one shot is 1 - 0.75 = 0.25.
To find the probability of not hitting the target in n shots, we multiply the probability of not hitting the target in one shot by itself n times (since the shots are independent events).
P(not hitting the target in n shots) = (0.25)^n
The probability of hitting the target at least once in n shots is the complement of not hitting the target in n shots, so:
P(hitting the target at least once in n shots) = 1 - P(not hitting the target in n shots) = 1 - (0.25)^n
We want this probability to be more than 0.99, so:
1 - (0.25)^n > 0.99
Solving this inequality for n:
0.01 > (0.25)^n
n * log(0.25) > log(0.01)
n * log(0.25) > log(0.01)
n > log(0.01) / log(0.25)
Using a calculator, we find:
n > -2 / -0.6021
n > 3.32
Since n must be a whole number, the minimum number of times the shooter must fire is 4.
Therefore, the correct answer is option A) 4.
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The probability of a shooter hitting a target is 0.75. How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?a)4b)2c)3d)1Correct answer is option 'A'. Can you explain this answer?
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