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VARVARA [1.3K]
1 year ago
7

Suppose that 40 batteries are shipped to an auto parts store, and that 4 of those are defective. A fleet manager then buys 8 of

the batteries from the store. In how many ways can at least 3 defective batteries be included in the purchase?
Mathematics
1 answer:
Assoli18 [71]1 year ago
7 0
From the given above, we will include the case in which 3 and all 4 defectives are included in the purchase. 

For 3:
                       (36C5) x (4C3) = 1507968
For 4:
                       (36C4) x (4C4) = 58905
Adding these numbers will give us an answer of 1566873.

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Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

Step-by-step explanation:

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P(N_1 = a) = \frac{5-a C 1}{5C2}

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P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}

And if we want to find the joint probability we just need to do this:

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P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

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