2) In a large vat of marbles, 30% are brown, 20% are red, 20% are yellow, 10% are green, 10% are orange, and the rest are blue. If a marble is drawn at random, what’s the probability that it’s (a) red or green (b) not yellow (c) neither yellow nor blue. If four marbles are drawn at random what’s the probability that (d) all are yellow (e) none are brown (f) exactly one is brown (g) at least one is brown (h) the first marble or the last marble is brown
Whats the formula for (f)? thanks!
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Anonymous2007-04-25 1:02 ID:cnID7IGy
This large vat I assume is large enough so that you can pretend that there are unlimited number of marbles? So then you could use binomial probability, where always P(brown) = 0.3 and P(not brown) = 0.7, no matter how many marbles you take out. So you want P(brown)xP(not brown)xP(not brown)xP(not brown).
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Anonymous2007-04-25 1:09 ID:cnID7IGy
Oh, and multiply the result above again by 4!/(1!(4-1!) which is 24/6=4. That's using the combinations formula, with 4C1.
The long way is to say you could have brown/notbrown/notbrown/notbrown OR notbrown/brown/notbrown/notbrown/ OR notbrown/notbrown/brown/notbrown OR notbrown/notbrown/notbrown/brown.
A certain automobile requires 12 (identical) chips all to function. 5% of the chips made by the manufacturer are defective, and when the automobile is made, 12 chips are independently chosen from this large shipment of chips. What’s the probability that the automobile will function?
also what formula would i use for this? thanks all you've been a lot of help so far!
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Anonymous2007-04-25 2:57 ID:7LKCYvlV
>>1
All answers are obvoiusly 7 (seven). Or 5 (five) if you like the law of fives.
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Anonymous2007-04-25 3:50 ID:n9AtPu95
a) P(R or G) = 3/10
b) P(not Y) = 8/10 = 4/5
c) P(not(Y or Bl)) = 7/10
d) P(ALL Y) = 2^4/715 = 16/715
e) P(NO Br) = 210/715 = 42/143
f) P(EXACTLY 1 Br) = 84/715
g) P(AT LEAST 1 Br) = 220/715 = 4/13
h) P(1st Br) = P(4th Br) = 660/715 = 12/13