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Sagot :
Using the hypergeometric distribution, it is found that there is a 0.1667 = 16.67% probability that he selects the two good batteries.
The batteries are chosen from the sample without replacement, which is why the hypergeometric distribution is used to solve this question.
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:
- There are 4 batteries, thus [tex]N = 4[/tex].
- Sample of 2 batteries, thus [tex]n = 2[/tex]
- 2 are good, thus [tex]k = 2[/tex].
The probability that both are good 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(2,4,2,2) = \frac{C_{2,2}C_{2,0}}{C_{4,2}} = 0.1667[/tex]
0.1667 = 16.67% probability that he selects the two good batteries.
A similar problem is given at https://brainly.com/question/24826394
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