Answer:
Step-by-step explanation:
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
is equal to
