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PROBABILITY QUESTIONS PLEASE HELP!!!?
1) If a car salesperson sells more than 3 cars during the week, the probability of taking the next weekend off is 7/8. If 3 or less cars are sold during the week, the probability of taking the next weekend off is 2/5. If the probability of selling more than 3 cars during the week is 1/5, find the probability that the car salesperson will take the next weekend off.
2) A bag contains n counters. Seven of these counters are green and the rest are yellow. Two counters are chosen at random. The probability that the two counters are green is 1/5.
a) Form an equation involving n and show that it simplifies to n^2 - n - 210 = 0
b) Find how many counters were in the bag originally
THANKS HELP WILL BE MUCH APPRECIATED!
3 Answers
- ?Lv 61 decade agoFavorite Answer
1)
W: He takes weekend off
A: He sells more than three cars
P[he takes weekend off] = P[W] = P[AWuA'W]
= P[AW] + P[A'W]
=P[A]*P[WIA] + P[A']*P[WIA']
=(1/5)*(7/8) + (4/5)*(2/5)
2)
7 green and (n-7) yellow
P[Selected counters are green] = 7C2/nC2 = 1/5 (given)
That is nC2 = 105
n(n-1) = 210
n^2 - n - 210 = 0
Solve and get n=15
- NoneLv 71 decade ago
1. This is conditional probability......
P(N > 3) = 1/5; IF N>3, P(weekend off) =7/8;
Probability that the car salesperson will take next weekend off = 1/5 x 7/8 = 7/40
2. P(1st counter green) = 7/n; P(2nd counter green) = 6/(n -1)
P(both counters green) = (7/n) x 6/(n-1) = 1/5.
42/(n^2 - n) = 1/5.........cross-multiply
210 = n^2 - n
n^2 - n - 210 = 0
(n - 15)(n +14) = 0
n = 15
- ignoramusLv 71 decade ago
1 ) P(selling 3 or more) = 1/5
P(weekend off given that sales ⥠3) = 7/8
P(both these events occur) = (1/5) x (7/8) = 7/40
P(selling < 3) = 4/5
P(weekend off given that sales < 3) = 2/5
P(both these events occur) = (4/5) x (2/5) = 8/25
Therefore P(he gets the weekend off) =
(7/40) + (8/25) = (35 + 64)/200 = 99/200. or 0.495
2 )
P(choosing a green) = 7/n
P(choosing a second green) = 6/(n-1)
Therefore P(choosing 2 green) = (7/n) x (6/(n - 1)) = 42/n(n - 1)
But if 42/n(n - 1) = 1/5
Then 5(42) = n(n - 1)
n² - n - 210 = 0
(n + 14) (n - 15) = 0
n = 15
There were 15 counters altogether.
