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The lifetime (in years) of bulbs is distributed as an Exp(1) random variable. Using Poisson approximation to the binomial distribution, the probability (round off to 2 decimal places) that out of the fifty randomly chosen bulbs at most one fails within one month equals
- a)0.05
- b)0.07
- c)0.09
- d)0.11
Correct answer is option 'C'. Can you explain this answer?
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The lifetime (in years) of bulbs is distributed as an Exp(1) random va...
**Given:**
- The lifetime of bulbs is distributed as an Exp(1) random variable.
- We need to find the probability that out of the fifty randomly chosen bulbs, at most one fails within one month.
**Approach:**
To calculate the probability, we will use the Poisson approximation to the binomial distribution.
**Poisson Approximation:**
If the number of trials (n) is large and the probability of success (p) is small, the binomial distribution can be approximated by the Poisson distribution with parameter λ = np.
In this case, the number of trials is 50 (n) and the probability of failure within one month is given by the exponential distribution with parameter λ = 1.
**Step 1: Calculate the probability of failure within one month:**
The exponential distribution with parameter λ = 1 is given by the probability density function:
f(x) = λ * e^(-λx)
To find the probability of failure within one month, we need to calculate the cumulative distribution function (CDF) at x = 1:
F(x) = 1 - e^(-λx)
Substituting the values, we get:
F(1) = 1 - e^(-1*1) = 1 - e^(-1) ≈ 0.632
Therefore, the probability of failure within one month is approximately 0.632.
**Step 2: Calculate the probability of at most one failure:**
To calculate the probability of at most one failure, we need to find the probability of 0 failures and the probability of 1 failure, and then sum them up.
The probability of 0 failures is given by:
P(X = 0) = e^(-λ) = e^(-1) ≈ 0.368
The probability of 1 failure is given by:
P(X = 1) = λ * e^(-λ) = 1 * e^(-1) ≈ 0.368
Therefore, the probability of at most one failure is approximately 0.368 + 0.368 = 0.736.
**Step 3: Subtract the probability of exactly one failure:**
Finally, we need to subtract the probability of exactly one failure from the probability of at most one failure.
P(at most one failure) - P(exactly one failure) = 0.736 - 0.368 = 0.368
Therefore, the probability that out of the fifty randomly chosen bulbs at most one fails within one month is approximately 0.368, which is closest to option 'C' (0.09).
- The lifetime of bulbs is distributed as an Exp(1) random variable.
- We need to find the probability that out of the fifty randomly chosen bulbs, at most one fails within one month.
**Approach:**
To calculate the probability, we will use the Poisson approximation to the binomial distribution.
**Poisson Approximation:**
If the number of trials (n) is large and the probability of success (p) is small, the binomial distribution can be approximated by the Poisson distribution with parameter λ = np.
In this case, the number of trials is 50 (n) and the probability of failure within one month is given by the exponential distribution with parameter λ = 1.
**Step 1: Calculate the probability of failure within one month:**
The exponential distribution with parameter λ = 1 is given by the probability density function:
f(x) = λ * e^(-λx)
To find the probability of failure within one month, we need to calculate the cumulative distribution function (CDF) at x = 1:
F(x) = 1 - e^(-λx)
Substituting the values, we get:
F(1) = 1 - e^(-1*1) = 1 - e^(-1) ≈ 0.632
Therefore, the probability of failure within one month is approximately 0.632.
**Step 2: Calculate the probability of at most one failure:**
To calculate the probability of at most one failure, we need to find the probability of 0 failures and the probability of 1 failure, and then sum them up.
The probability of 0 failures is given by:
P(X = 0) = e^(-λ) = e^(-1) ≈ 0.368
The probability of 1 failure is given by:
P(X = 1) = λ * e^(-λ) = 1 * e^(-1) ≈ 0.368
Therefore, the probability of at most one failure is approximately 0.368 + 0.368 = 0.736.
**Step 3: Subtract the probability of exactly one failure:**
Finally, we need to subtract the probability of exactly one failure from the probability of at most one failure.
P(at most one failure) - P(exactly one failure) = 0.736 - 0.368 = 0.368
Therefore, the probability that out of the fifty randomly chosen bulbs at most one fails within one month is approximately 0.368, which is closest to option 'C' (0.09).
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The lifetime (in years) of bulbs is distributed as an Exp(1) random variable. Using Poisson approximation to the binomial distribution, the probability (round off to 2 decimal places) that out of the fifty randomly chosen bulbs at most one fails within one month equalsa)0.05b)0.07c)0.09d)0.11Correct answer is option 'C'. Can you explain this answer?
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