Question 870692: I need help as soon as possible! ANY HELP at all is really appreciated. I am completely stuck....
(1) Exactly the half of the 10 batteries, that you save on a closet, are not charged but you don't know which of them are. You are in a hurry to leave on holidays and you take 4 batteries from the closet but you don't have the time to check if they are not charged or charged. What is the probability none of the 4 batteries is charged?
(2) Before you leave on holidays, you remember that in the drawer of your desk, there are 10 other batteries, but again the half of them are not charged and you don't know which of them are, neither you have the time to check them. Due to the space on your bag, you can ONLY take 4 batteries for your trip.
Compare the below options towards the probability that all the batteries that you took are not charged.
(i) You take 4 batteries from the drawer and none of them from the closet.
(ii) You take 2 batteries from the closet and 2 from the drawer.
(iii) You put all 20 batteries in a bag, you mix them up and you pick 4, and you take them.
Found 2 solutions by ewatrrr, rothauserc:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! You can feel comfortable using n = 10
Might recommend stattrek.com(binomial P) for a good reference and check to Your work.
Note: Using n = 10
Yes, n = 10 (taking from a closet/drawer with 10 batteries in them)
AND the probability referred to those 10 alone. (Not in general, so to speak)
1) closet: p(not charged) = .5, n =10,
P(x = 4) = 10C4(.5^4)(.5^6) = 210(.5^4)(.5^6) = .20
2) drawer: p(not charged) = .5
i) n = 10, P(x = 4)= 10C4(.5^4)(.5^6) = 210(.5^4)(.5^6) = .20
ii)n = 10, P(x =2) = 10C2(.5^2)(.5^8) = .0439
.0439 + .0439 = .0878
iii)p = .5, n = 20, P(x = 4) =20C4(.5^4)(.5^16) = .0046
3) Compare the options of the previous question (2) towards the mean number of charged batteries that you took.
As p(charged) = .5,
the computations as to the various P(charged batteries) would have same results
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! 1) there are 5 batteries that are not charged
probability that the 4 batteries picked are not charged = 5/10 * 4/9 * 3/8 * 2/7 = 120/5040 = 1/42 is approximately 0.02
2)
i) same as 1
ii) 5/10 * 4/9 * 5/10 * 4/9 = 400/8100 is approximately 0.05
iii) 10/20 * 9/19 * 8/18 * 7/17 = 5040/116280 is approximately 0.04
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