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If the probability of choosing 2 red marbles without replac
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23 May 2017, 08:49
23 May 2017, 08:49
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65% (01:52) correct
34% (02:06) wrong based on 285 sessions
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If the probability of choosing 2 red marbles without replacement from a bag of only red and blue marbles is \(\frac{3}{55}\) and there are 3 red marbles in the bag, what is the total number of marbles in the bag?
A) 10
B) 11
C) 55
D) 110
E) 165
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Re: If the probability of choosing 2 red marbles without replac
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24 May 2017, 12:55
24 May 2017, 12:55
9
Carcass wrote:
If the probability of choosing 2 red marbles without replacement from a bag of only red and blue marbles is \(\frac{3}{55}\) and there are 3 red marbles in the bag, what is the total number of marbles in the bag?
A) 10
B) 11
C) 55
D) 110
E) 165
Let T = the TOTAL number of marbles in the bag.
We are told that 3 of those marbles are red
P(select a red marble first) = 3/T
Once we've removed the first red marble, there are T-1 marbles remaining and 2 of them are red
So, P(select a red marble second) = 2/(T-1)
Okay, now let's use probability rules to answer the question....
P(select 2 red marbles) = P(select a red marble first AND select a red marble second)
= P(select a red marble first) x P(select a red marble second)
= 3/T x 2/(T-1)
= 6/(T² - T)
We're told that P(select 2 red marbles) = 3/55
So, we can write: 6/(T² - T) = 3/55
Cross multiply to get: 3(T² - T) = (6)(55)
Divide both sides by 3 to get: T² - T = (2)(55)
Evaluate: T² - T = 110
Rearrange: T² - T - 110 = 0
Factor: (T - 11)(T + 10) = 0
So, EITHER T = 11 OR T = -10
Since T cannot be negative, it must be the case that T = 11
Answer:
Show: ::
B
Cheers,
Brent
General Discussion
Re: If the probability of choosing 2 red marbles without replac
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03 Feb 2018, 21:54
03 Feb 2018, 21:54
2
An easy approach :
Let no: of blue bags be B. Given no: of red bags is 3. Thus
3/(3+B) * 2/(3+B-1) = 3/55 (Probablity without replacement formula)
We need to know 3+B
On solving,
(3+B)(3+B-1)=110
Thus 3+B is 11
Let no: of blue bags be B. Given no: of red bags is 3. Thus
3/(3+B) * 2/(3+B-1) = 3/55 (Probablity without replacement formula)
We need to know 3+B
On solving,
(3+B)(3+B-1)=110
Thus 3+B is 11
