Using the hypergeometric distribution, it is found that there is a 0.0455 = 4.55% probability that both light bulbs failed the test.
The bulbs are chosen without replacement, and this is why the hypergeometric distribution is used to find the probability.
Hypergeometric distribution:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- 12 bulbs, thus [tex]N = 12[/tex]
- 3 failed the test, thus [tex]k = 3[/tex].
- 2 are selected, thus [tex]n = 2[/tex].
The probability that both failed the test is P(X = 2), thus:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 2) = h(2,12,2,3) = \frac{C_{3,2}C_{9,0}}{C_{12,2}} = 0.0455[/tex]
0.0455 = 4.55% probability that both light bulbs failed the test.
A similar problem is given at https://brainly.com/question/24826394
