There are 20 different beads, 4 of which are yellow. Choosing 30 beads with repetitions, I need to calculate:
- The probability that exactly 5 of the chosen beads are yellow
- The probability that the first and last beads are yellow and exactly 5 yellow beads were chosen (including the first and last beads)
The first one I calculated saying there are $4^5$ ways to chose 5 yellow beads and $16^{25}$ ways to choose the rest hence $$P_1 = \frac{4^5\cdot 16^{25}}{20^{30}}=\frac{4^{55}}{4^{30}\cdot5^{30}}=(0.8)^{25}\cdot(0.2)^{5}$$ As for the second probability I first tried choosing $\binom{5}{2}$ yellow beads for first and last and then multiply by 2 as they can change places and again multiply by the permutation of the remaining beads $28!$ hence $$P_2 = P_1\cdot\frac{\binom{5}{2}\cdot2\cdot28!}{30!}$$ but this is incorrect because there are repetitions. So the second attempt was choosing first and last yellow bead which can be done in $5^2$ and then $$P_2 = P_1\cdot\frac{5^2\cdot28!}{30!}$$ but this seems wrong too. My intuition says that the probability between 1 and 2 remains the same but I haven't found a way to prove (or disprove) it
Best Answer
Because we are asked about probability, and not about the number of combinations, and because we are choosing with repetitions - we can ignore the identity of the individual beads and instead treat each choice as a $\frac{4}{20} = 20\%$ probability to choose yellow and $80\%$ probability to choose non-yellow.
For any given choice of the 5 places where yellow beads are chosen, the probability to choose it is $$ 0.2^5\cdot 0.8^{25} = \frac{2^5\cdot 8^{25}}{10^5\cdot 10^{25}} = \frac{2^5\cdot 2^{75}}{10^5\cdot 10^{25}} = \frac{2^{80}}{10^{30}} $$
Since each such case is disjoint to the others, we can calculate how many ways there are to choose these 5 places and multiply: $$\boxed{ \frac{2^{80}}{10^{30}} \cdot \binom{30}{5} = \frac{2^{80}}{10^{30}} \cdot \frac{30!}{5!\cdot 25!} \approx 17.2279\% }$$
For the second, the first and last places always have yellow beads so we only choose 3 places out of 28: $$\boxed{ \frac{2^{80}}{10^{30}} \cdot \binom{28}{3} = \frac{2^{80}}{10^{30}} \cdot \frac{28!}{3!\cdot 25!} \approx 0.396\% }$$