Suppose that a box contains 6 cameras and that 4 of them are defective. a sample of 2 cameras is selected at random. define the random variable x as the number of defective cameras in the sample. ( round to two decimal places when needed.)

Since we sample two cameras, the random variable X can only have three possible values:

0 - both cameras work correctly

1 - only one defective camera

2 - both cameras are defective.

To fully define the random variable, we need to compute the proability for each of the possible values.

X follows a Bernoulli distribution, where we want to count the number of "successes" (finding a defective camera) on a sample of 6 elements. Since 4 out of the 6 cameras are defective, the probability of finding a defective camera is 4/6=2/3.

In general, we have

So, the probabilities are

the answer for your question is

step-by-step explanation:

27/4x^6 y^8

step-by-step explanation:

4(3x^2y^4)^3 (2x^3y^5)^4

=

108x^6 y^12 /16x^12 y^20

=

108 /16x^6 y^8

=

27 /4x^6 y^8