among 10 laptop 5 are good and 5 have defects .unaware of this a customer buys 6 laptops

(a) what is the probability that exactly two of them are defective

(b)given that atleast 2 purchases laptops are defective,what is the probability that exactly 2 are defective?

Question

among 10 laptop 5 are good and 5 have defects .unaware of this a customer buys 6 laptops

(a) what is the probability that exactly two of them are defective


(b)given that atleast 2 purchases laptops are defective,what is the probability that exactly 2 are defective?

anonymous · Answer

Take note of the formula for the binomial probability.

anonymous · Answer

$$\displaystyle P(x) = \frac{n!}{x!*(n-x)!}*p^x*(1-p)^{n-x}$$

anonymous · Answer

where n is the sample size, p is the probability of success, and x is the number of successes.

anonymous · Answer

The factorial works like 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800..