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:

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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