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
(A) 0.001 (B) 0.057 (C) 0.107 (D) 0.3
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Option D
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Answer:
Probability of 3 defective out of 10 is 0.057
Hence answer is B
Step-by-step explanation:
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 ∗ p^x ∗ 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 ∗ p^3 ∗ q^(10−3)
= 120 ∗ 0.1^3 ∗ 0.9^7
= 0.057
Hence answer is B
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