SOLUTION: A box contains 6 pink,7 blue,4 white and 5 red bulbs.if 4 bulbs drawn at random,determine the probability that, (a) all four are blue (b) two are pink and two are white (c) at l

Question 253207: A box contains 6 pink,7 blue,4 white and 5 red bulbs.if 4 bulbs drawn at random,determine the probability that,
(a) all four are blue
(b) two are pink and two are white
(c) at least one is red

Found 2 solutions by drk, Greenfinch:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
I assume that you do not replace. These are combination problems:
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(a) p (4 blue) = 6c0 * 7c4 * 4c0 * 5c0 / 22c4 = 35 / 7315 = .00478
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(b) P(2 pink and 2 white): 6c2 * 7c0 * 4c2 * 5c0 / 22c4 = 90 / 7315 = .0123
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(c) p(at least one is red). We don't want any red, so we choose all the others.
P(at least 1 red) = 17c4 * 5c0 / 22c4 = 2380 / 7315 = .3254
Answer by Greenfinch(383) (Show Source): You can put this solution on YOUR website!
This type of problem divides into two types. Dependent, where the first pick affects the second and so on. ie You take a candy from the bag, note the colour and then EAT it. The bag contents are now different from the first pick and so the chances are different. The second is Independent, when you put the candy back and the bag contents are the same as the first pick (and you are rather weird)
a) This is dependent, so, initially there are 22 bulbs or which 7 are blue so chance is 7/22. Next pick there are 6 out of 21, then 5 of 20, and finally 4 of 19
Chance is (7/22) x (6/21) x (5/20) x (4/19) = 1/209
b) This is a combination question because the order is not important as long as you end up with two of each colour. The possibilities are
PPWW
PWPW
PWWP
WPPW
WPWP
WWPP
So ,using the technique in a)
PPWW = (6/22) x (5/21) x (4/20) x (3/19) = 3/1463
PWPW = (6/22) x (4/21) x (5/20) x (3/19) = 3/1463
PWWP = (6/22) x (4/21) x (3/20) x (5/19) = 3/1463 and so on...
so it doesn't matter about the order, the chances are the same, the number of ways we can achieve this is 6.
So chance is 6x(3/1463) = 18/1463
c)At least one is red. Easiest way is to work out the chance of no reds and subtract from 1. Or you can work out all the chances of 1, 2, 3 or 4 reds
Since there are 5 reds, the chance of a non red on first pick is 17/22
The second pick will still have 5 reds, but 21 total, so 16/21
So chance (17/22) x (16/21) x (15/20) x (14/19) = 68/209 of no reds
chance of at least red is 1 - (68/209)= 141/209
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