Question

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

Answer 1

Here's your answer

Option D

Answer 2

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