I am not good with these combination questions but was wondering if someone could help me figure out this question:
A bag contains:
5 blue beads
6 red beads
9 green beads
20 yellow beads
so 40 beads in total.
A person draws the beads in succession until they get 5 matching colored beads (order does not matter). For instance, if they draw 1 red, then 1 blue, 1 red, 1 red, 1 green, 1 red, 1 red then they have got 5 matching reds so the game finishes. They must continue to draw until 5 matching colours are made.
I am wondering how to calculate the probability that they draw 5 blue beads or 5 red beads. I assume I can calculate the probability of each separately then add up?
Now does anyone have any ideas how to calculate that probability for 5 blue?
I am not sure how to deal with the fact that it has to be 5 blue balls with none of the others being at 5.
Best Answer
Let's say $p(i,j,k)$ is the probability that the 5th blue bead is drawn immediately after drawing $i$ red beads, $j$ green beads, and $k$ yellow beads where $0 \le i,j,k \le 4$. After the previous draw, just before drawing the 5th blue bead, we must have drawn exactly $4$ blue beads; then, with $36-i-j-k$ beads remaining in the bag, we must draw the final blue bead. So $$p(i,j,k) = \frac{\binom{5}{4} \binom{6}{i} \binom{9}{j} \binom{20}{k}}{\binom{40}{4+i+j+k}} \cdot \frac{1}{36-i-j-k}$$
The probability that 5 blue beads are drawn before 5 of any other color is then $$ \sum_{i=0}^4 \sum_{j=0}^4 \sum_{k=0}^4 p(i,j,k) = 0.00194347 $$
A similar computation for the case of drawing 5 red beads before 5 of any other color yields a probability of $ 0.00824479$. So the probability drawing 5 blue or 5 red beads before 5 of any other color is $0.00194347 + 0.00824479 = 0.0101883$.