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The probability that a thermistor randomly picked up from a production unit is defective is 0.1. The probability that out of 10 thermistors randomly picked up, 3 are defective is (2015)
- a)0.001
- b)0.057
- c)0.0107
- d)0.3
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The probability that a thermistor randomly picked up from a production...
Let p (getting a defective thermistor ) = p = 0.1
Let q (getting a working thermistor ) q = 1 − p = 0.9
Suppose X has a binomial distribution, the probability of x success in n-Bernoulli trials,
Here in this case getting a defective thermistor is success,
p(X = x) = nCx ∗ px ∗ q(n–x)
where in x = 0, 1, 2,..., n some finite number of required successes out of some finite number of trials(n)
Here n = 10 and x = 3
P (3 of thermistors are defective) = p(X = 3)
= 10C3 ∗ p3 ∗ q(10−3)
= 120 ∗ 0.13 ∗ 0.97
= 0.057
Most Upvoted Answer
The probability that a thermistor randomly picked up from a production...
Let p (getting a defective thermistor ) = p = 0.1
Let q (getting a working thermistor ) q = 1 − p = 0.9
Suppose X has a binomial distribution, the probability of x success in n-Bernoulli trials,
Here in this case getting a defective thermistor is success,
p(X = x) = nCx ∗ px ∗ q(n–x)
where in x = 0, 1, 2,..., n some finite number of required successes out of some finite number of trials(n)
Here n = 10 and x = 3
P (3 of thermistors are defective) = p(X = 3)
= 10C3 ∗ p3 ∗ q(10−3)
= 120 ∗ 0.13 ∗ 0.97
= 0.057
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Community Answer
The probability that a thermistor randomly picked up from a production...
Calculation:
To calculate the probability that out of 10 thermistors randomly picked up, 3 are defective, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- n = total number of trials (10 thermistors)
- k = number of successful trials (3 defective thermistors)
- p = probability of success (0.1, probability of a defective thermistor)
Plug in the values:
P(X = 3) = (10 choose 3) * 0.1^3 * (1-0.1)^(10-3)
P(X = 3) = (10! / (3! * (10-3)!)) * 0.001 * 0.387420489
P(X = 3) = 120 * 0.001 * 0.387420489
P(X = 3) = 0.04649
Therefore, the probability that out of 10 thermistors randomly picked up, 3 are defective is approximately 0.04649 or 0.057 (rounded to three decimal places).
To calculate the probability that out of 10 thermistors randomly picked up, 3 are defective, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- n = total number of trials (10 thermistors)
- k = number of successful trials (3 defective thermistors)
- p = probability of success (0.1, probability of a defective thermistor)
Plug in the values:
P(X = 3) = (10 choose 3) * 0.1^3 * (1-0.1)^(10-3)
P(X = 3) = (10! / (3! * (10-3)!)) * 0.001 * 0.387420489
P(X = 3) = 120 * 0.001 * 0.387420489
P(X = 3) = 0.04649
Therefore, the probability that out of 10 thermistors randomly picked up, 3 are defective is approximately 0.04649 or 0.057 (rounded to three decimal places).
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The probability that a thermistor randomly picked up from a production unit is defective is 0.1. The probability that out of 10 thermistors randomly picked up, 3 are defective is (2015)a)0.001b)0.057c)0.0107d)0.3Correct answer is option 'B'. Can you explain this answer?
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