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grin007 [14]
2 years ago
5

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

at 3 or less (P (x <= 3) are blemished?
Mathematics
1 answer:
hodyreva [135]2 years ago
6 0

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in ^{5000}C_{10} ways.

Possibility 1: <u>3 tires are blemished.</u>

Total 10 tires can be chosen with 3 blemished tires in ^{1000}C_3 \times {4000}C_7 ways.

Possibility 2:  <u>2 tires are blemished.</u>

Total 10 tires can be chosen with 2 blemished tires in ^{1000}C_2 \times ^{4000}C_8  ways.

Possibility 3: <u>1 tires are blemished.</u>

Total 10 tires can be chosen with 1 blemished tires in ^{1000}C_1 \times^{4000}C_9 ways.

Possibility 4:  <u>No tires are blemished.</u>

Total 10 tires can be chosen with no blemished tires in ^{1000}C_0 \times^{4000}C_{10} ways.

Hence, the required probability is \frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }.

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