Answer:
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.
