Question

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.)

Answer 1

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

$P(X = k) = \binom{2}{k} \left(\cfrac{2}{3}\right)^k\left(\cfrac{1}{3}\right)^{2-k}$

So, the probabilities are

$P(X = 0) = \binom{2}{0} \left(\cfrac{2}{3}\right)^0\left(\cfrac{1}{3}\right)^{2} = \cfrac{1}{9} = 0.\overline{1} \approx 0.11$

$P(X = 1) = \binom{2}{1} \left(\cfrac{2}{3}\right)\left(\cfrac{1}{3}\right) = \cfrac{4}{9} = 0.\overline{4} \approx 0.44$

$P(X = 2) = \binom{2}{2} \left(\cfrac{2}{3}\right)^2\left(\cfrac{1}{3}\right)^{0} = \cfrac{4}{9} = 0.\overline{4} \approx 0.44$