probability of Marble randomly picked

Question

A bag contains three red marbles, two green ones, one pearl blue two yellow and three orange marbles. if you grab at random 5 marbles what is the probability that you have all red ones and the pearl blue marble?

Answer 1

The probability of grabbing 3 red marbles and 1 pearl blue marble out of the bag when grabbing 5 total marbles is equal to (the number of possible combinations of 5 marbles that include 3 red marbles and 1 pearl blue marble), divided by (the total number of possible combinations of 5 marbles).

How many different possible combinations of 5 marbles are there when grabbing out of a bag of 11 marbles (assuming the order of marbles does not matter)? We can use the Combinations formula.

C(n,k) = n!/(k!*(n-k)!)

n = 11 marbles, k = 5 marbles

C(11,5) = 11!/(5!*(11-5)!) = 462

So there are 462 unique handfuls of marbles that we can grab from the bag, each of which is equally likely.

How many different ways are there to grab 5 marbles such that 3 are red, and 1 is pearl blue (assuming the order of marbles does not matter)? Note that there are 7 possible ways, because other than the 3 red and 1 pearl blue marble, there are 7 other marbles, one which must be the 5th marble of our set of 5 marbles. I.e. you could grab...

Red1-Red2-Red3-Blue1-Green1

Red1-Red2-Red3-Blue1-Green2

Red1-Red2-Red3-Blue1-Yellow1

Red1-Red2-Red3-Blue1-Yellow2

Red1-Red2-Red3-Blue1-Orange1

Red1-Red2-Red3-Blue1-Orange2

Red1-Red2-Red3-Blue1-Orange3

So there are 7 possible handfuls of marbles that satisfy our requirements of 3 red marbles and 1 blue marble. These 7 possibilities are included within the 462 total possibilities that we calculated, where each possibility is equally likely of occurring, so the probability of one of these 7 possibilities occurring is:

Probability = 7/462 = .01515..., or about 1.5%.

Answer 2

The most common way to brute force the probability is to find the number of ways 3R, 1B, and 1 other can be obtained divided by all the ways 5 marbles can be drawn from a group of 11 marbles.

(The ways to draw 3R out of 3R)(The ways to draw 1B out of 1B)(The ways to draw 1 other out of 7 others)/(The ways to draw 5 balls out of 11)

Formally: ₃C₃ * ₁C₁ * ₇C₁ / ₁₁C₅ = 7 / (11!/(6! 5!)) = 7/462