Probablity?

Question

A jar contains 5 red marbles and 4 yellow marbles. Four marbles are chosen without replacement. What is the probability of picking at least 3 red marbles? Can anyone show me how you do this problem...it would be greatly appreciated.

Answer 1

The number of red marbles in the marbles that are selected has a hypergeometric distribution with

Number in population = N = 9 (5+4)
Number of successes in population = k = 5
Number of draws = n = 4.

The probability distribution is

P(X = x) = (kCx) * {(N-k)C(n-x)} / (NCn)

BTW, kCx = k!/{x!(k-x)!}.

So if we want to find the probability of at least 3 successes, then

P(X >= 3) = P(X = 3) + P(X = 4)

= (5C3) * (4C1) / (9C4) + (5C4) * (4C0) / (9C4)

= 40/126 + 5/126

= 45/126

edit: jacobandalex, the draws are made without replacement, and there are four draws, not three. Your answer is incorrect. It is easiest to use the hypergeometric distribution for this problem.

Answer 2

We have the following possible chains of four outcomes leading to at least three red marbles (R) chosen. all other chains lead to failing to select at least three R's:

RRRR = 5/9 4/8 3/7 2/6 = ? multiple of the four factors
RRRY = 5/9 4/8 3/7 4/6 = ? ditto
RRYR = 5/9 4/8 4/7 3/6 = ? ditto
RYRR = 5/9 4/8 4/7 3/6 = ? ditto
YRRR = 4/9 5/8 4/7 3/6 = ? ditto

Thus, P(nR >= 3) = the sum of the ?'s indicated above. You can do the arithmatic if you need a real number.

All the chains, including ones that do not end with at least three R's, can be modeled as a binomial expansion of (R + Y)^4 = R^4 + 4R^3 Y + 6R^2 Y^2 + 4R Y^3 + Y^4 where only the first two terms lead to at least four R's picked. The second term verifies there are four ways to select three R's and one Y. The first term shows the unlikely case where we select four R's.

In general, we might use the binomial expansion to calculate the probability of getting at least three R's, but if and only if there were replacement after each draw so that P(R) = p and P(Y) = q = 1 - p for all draws. However, in your case, we have to account for the non replacement. Which means we need to account for the order in which each R or Y is selected.

In that first chain, for example, it is clear there are 5 R's and 4 Y's in the pot; so chances are 5/9 an R will be selected. But now, with one R gone, the chances are 4/8 the second draw will get an R.

Which is one reason why I chose to solve this problem using event chain analysis. The other is that by drawing out the chains, as I've done above, I can actually visualize what affects the final probability result.

Answer 3

The equation is:
# of favourable outcomes
__________________
# of total outcomes

The probability of two things happening consecutively, the probabilities are multiplied:

5/9 * 5/9 * 5/9 =
5*5*5
____
9*9*9

The answer is 125/729