Find the probability distribution of random variable

Reference no: EM13225612

A very large shipment of parts contains 10% defectives. 2 parts are chosen at random from the shipment and checked. Let the random variable X denote the number of defectives found. Find the probability distribution of this random variable.

SOLUTION:
P(0) = (.90)(.90)=.81
P(1) = (.90)(.10)+(.10)(.90)=.18
P(2)=(.10)(.10) = .01

This is the solution that I completed earlier this summer and was correct. NOW I can't figure out how I figured that out! why is p(0)=(.9)(.9) but p(1)=(.9)(.1)+(.1)(.9). I don't understand why I added the numbers in the order that I did for p(1) and I don't understand why the other probabilities aren't the same. Please provide me with the logic on why it is multiplied with these number and added in the order that it is please!

Reference no: EM13225612

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