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Answer:
The probability that a box with two defective toys will be shipped is 0.29825
Step-by-step explanation:
The first toy picked has a 17 in 19 chance of not being defective
The second toy picked has a 16 in 18 chance of not being defective
The third toy picked has a 15 in 17 chance of not being defective
The probability that the toy is not defective, and will be shipped
P(toy not defective and shipped) = (17/19) *(16/18)*(15/17)
= (16*15)/(19*18)
= 240/342
= 0.70175
The question said, suppose that a given box has two defective toys, what is the probability that it will be shipped?
P( toy defective and not shipped) = 1 - P(toy not defective and shipped)
= 1-0.70175
= 0.29825
The probability that a box with two defective toys will be shipped is 0.29825
Using the hypergeometric distribution, it is found that there is a 0.7018 = 70.18% probability that it will be shipped.
The toys are chosen without replacement, which is the reason why the hypergeometric distribution is used.
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:
- 19 toys, thus [tex]N = 19[/tex].
- 2 are defective, thus [tex]k = 2[/tex]
- 3 are selected, thus [tex]n = 3[/tex].
It will be shipped if there are no defectives, thus the probability is P(X = 0), then:
[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 = 0) = h(0,19,3,2) = \frac{C_{2,0}C_{17,3}}{C_{19,3}} = 0.7018[/tex]
0.7018 = 70.18% probability that it will be shipped.
A similar problem is given at https://brainly.com/question/24826394
