A corporation has 11 manufacturing plants. Of these, 7 are domestic and 4 are located outside the United States. Each year a performance evaluation is conducted for 4 randomly selected plants.

a. What is the probability that a performance evaluation will include 2 or more plants from outside the United States?

b. What is the probability that a performance evaluation will contain 3 plants from the United States?

c. What is the probability that a performance evaluation will include exactly 1 plant outside the United States?

Question

Rosemary Norris

Mathematics

3 August, 23:28

A corporation has 11 manufacturing plants. Of these, 7 are domestic and 4 are located outside the United States. Each year a performance evaluation is conducted for 4 randomly selected plants.

a. What is the probability that a performance evaluation will include 2 or more plants from outside the United States?

b. What is the probability that a performance evaluation will contain 3 plants from the United States?

c. What is the probability that a performance evaluation will include exactly 1 plant outside the United States?

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Answers (1)

Know the Answer?

Belinda Osborne · Answer 1

a) The probability that 2 or more plants are from outside the Us is 0.4697

b) If with contain 3 you mean 3 or more, then the probability that the performance contain 3 plants from the united states would be 0.5303. If with contain 3 you mean exactly 3, then the answer is the same as item (c), 0.4242

c) The probability that exactly 1 plant is from outside the Us is 0.4242.

Step-by-step explanation:

a) I will compute the probability of the complementary event, that is, that the proformance evaluation include 1 or none plants outside the Us. Then i will substract that probability from 1.

The probability that all 4 plants are from the Us is 7/11*6/10*5/9*4/8 = 7/66 = 0.10606

The probability that 3 plants are from the Us is

7/11*6/10*5/9*4/8 * 4 = 14/33 = 0.4242 (note that we can compute the probability that the first 3 plants are from Us and the fourth one from outside and multiply that by 4, the total amount of ways of ordering the outsider plant).

The probability that 1 or less plants are from outside the Us is, as a consecuence, 7/66 + 14/33 = 35/66 = 0.5303.

The probability that 2 or more plants are from outside the Us is 1-0.5303 = 0.4697

b) i will assume that in the event 'the permormance evaluation contains 3 plants from the Us', 4 plants from the Us is also a possibility in this event, otherwise this event would be equal from that one of item (c) (They would express the same with different words).

Now, the probability of getting 3 or more plants from the Us was alredy computed: it is the probability that 1 or less plants are from outsude the Us, 0.5303

c) The probability that exactly one plant is from outside the Us is equal to the probability that exactly 3 plants are from the Us (is an equivalent event). Thus, this probability was alredy computed and it is 0.4242.