There're 5 white balls, 9 black balls, and 14 red balls. What's the probability that if I pick out 6 balls without replacement, it will be 3 white, 2 black and a red ball?
Help appreciated!
I think the answer is (28 choose 6)/3! 2!, is that correct?
[Math] Choosing colored balls
combinatorics
Related Solutions
There are $150$ balls altogether, and you’re choosing $20$ of them; a set with $150$ members has $\binom{150}{20}$ $20$-element subsets, so there are $\binom{150}{20}$ sets of $20$ balls that you could draw, all of them equally likely to be drawn. That accounts for the denominator of the fraction: it’s the number of equally likely possible outcomes.
Similarly, there are $\binom{40}{10}$ different sets of $10$ white balls, $\binom{50}4$ different sets of $4$ red balls and $\binom{60}6$ different sets of $60$ black balls. There are $$\binom{40}{10}\binom{50}4\binom{60}6$$ ways to combine one of the $\binom{40}{10}$ possible sets of $10$ white balls, one of the $\binom{50}4$ possible sets of $4$ red balls, and one of the $\binom{60}6$ possible sets of $6$ black balls, so there are $$\binom{40}{10}\binom{50}4\binom{60}6$$ successful outcomes, where successful means having $10$ white balls, $4$ red balls, and $6$ black balls. As usual, the probability of a successful outcome is the ratio of successful outcomes to equally likely possible outcomes, or
$$\frac{\binom{40}{10}\binom{50}4\binom{60}6}{\binom{150}{20}}\;.$$
Nothing in the analysis changes in any way if we paint numbers from $1$ through $150$ on the balls to give them unique identities: we still have to count the number of sets of $10$ white balls, the number of sets of $4$ red balls, the number of sets of $6$ black balls, and the number of sets of $20$ balls of whatever colors, and we still have to perform the same calculations with these numbers.
Added: I’m afraid that the calculation in the edit makes little sense. To see this more clearly, let’s look at a simpler example. Suppose that there are $99$ white balls and $1$ red ball, and you draw $20$ balls at random without replacement. By your reasoning there are $\binom{21}1=21$ possible outcomes, ranging from $20$ white and no red balls to no white and $20$ red balls. This, however, is clearly not the case, since there is only $1$ red ball in the bag. Okay, suppose that you take this limitation into account: then by your approach there are exactly two possible outcomes: $20$ white balls, and $19$ white balls and $1$ red balls, so the denominator of your fraction will be $2$.
For the outcome of drawing $20$ white balls the numerator will be $\frac{20!}{20!}=1$, so you would conclude that the probability of getting $20$ white balls is $\frac12$. Is that reasonable? Notice that you’d get the same result if there were $999$ white balls and $1$ red ball, or $999999$ white balls and $1$ red ball. I think that it’s pretty clear that this cannot be right.
Worse, your numerator for the outcome of drawing $19$ white balls and the red ball will be $\frac{20!}{1!19!}=20$, and the ‘probability’ of this outcome will be $\frac{20}2=10$. That certainly can’t be right!
What you’re missing is that even if there is nothing to distinguish one white ball from another, there are still $99$ different white balls in the bag, and a $99$-element set has $\binom{99}{20}$ different $20$-element subsets. Every one of these $\binom{99}{20}$ sets is a different outcome, even if you have no way to tell which of them you actually got. Thus, in the simplified problem the probability of getting $20$ white marbles is actually
$$\frac{\binom{99}{20}}{\binom{100}{20}}=\frac{99!}{20!79!}\cdot\frac{20!80!}{100!}=\frac{80}{100}=\frac45\;,$$
while the probability of getting $19$ white balls and the red ball is simply the probability of getting the red ball, $$\frac{20}{100}=\frac15\;.$$
We can consider each configuration for without replacement as a sequence of balls. Then permuting the sequence in any way gives another configuration with equal probability because every sequence has equal probability. Each permutation can be used to give a bijection (one-to-one correspondence) between a set of configurations with another that differ simply by that permutation.
For example, in the first question you want #(sequences with 5th ball red)/#(sequences), and #(sequences with 5th ball red) = #(sequences with 1st ball red) by the bijection that switches the 1st and 5th ball in a sequence. Check that every sequence with 5th ball red corresponds to exactly one sequence with 1st ball red, and vice versa. This gives the answer since #(sequences with 1st ball red)/#(sequences) = 3/9 because there are no restrictions on the later balls.
For the second question, to do it rigorously we choose some permutation that sends the 3rd ball to position 1, the 7th ball to position 2 and the 9th ball to position 3. Then we get #(sequences with 3rd and 7th ball red and 9th ball black) = #(sequences with 1st and 2nd ball red and 3rd ball black), and #(sequences with 3rd and 7th ball red) = #(sequences with 1st and 2nd ball red). So the answer is #(sequences with 1st and 2nd ball red and 3rd ball black)/#(sequences with 1st and 2nd ball red) = 2/7 since the first two balls already use up two red and after drawing the first two we are left with 7 balls of which 2 are black.
Best Answer
HINT: There are $\binom{28}6$ different ways to pick $6$ balls, and they’re equally likely. If $n$ is the number of ways to pick a combination containing $3$ white balls, $2$ black balls, and one red ball, then the probability of picking such a combination is $$\frac{n}{\binom{28}6}\;:$$ the total number of possibilities goes in the denominator, not the numerator.
To calculate $n$, observe that there are $\binom53$ ways to choose $3$ white balls from the $5$ that are available.
How many ways are there to pick $2$ black balls from a set of $9$?
How many ways are there to pick one red ball from a set of $14$?
How should you combine these various numbers to get $n$, the number of successful combinations?