1
$\begingroup$

There are 20 different beads, 4 of which are yellow. Choosing 30 beads with repetitions, I need to calculate:

  1. The probability that exactly 5 of the chosen beads are yellow
  2. 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

$\endgroup$

2 Answers 2

2
$\begingroup$

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: $$\boxed{ \frac{2^{80}}{10^{30}} \cdot \binom{28}{3} = \frac{2^{80}}{10^{30}} \cdot \frac{28!}{3!\cdot 25!} \approx 0.396\% }$$

$\endgroup$
3
  • 1
    $\begingroup$ There are 30 draws, not 20. $\endgroup$ Oct 20, 2020 at 13:03
  • $\begingroup$ @JaapScherphuis It was 20 before OP edited the question. Fixed. $\endgroup$
    – Idan Arye
    Oct 20, 2020 at 13:52
  • $\begingroup$ Sorry, I hadn't noticed the question had been edited. $\endgroup$ Oct 20, 2020 at 14:12
1
$\begingroup$

I disagree with the posted solution.

  1. it is a Binomial $X\sim B(30;\frac{4}{20})$

Thus the solution is

$$\mathbb{P}[X=5]=\binom{30}{5}\Big(\frac{4}{20}\Big)^5\Big(\frac{16}{20}\Big)^{25}\approx 17.23\%$$

2.

$$\mathbb{P}[X=5, \text{ first and last Yellow}]=\binom{28}{3}\Big(\frac{4}{20}\Big)^5\Big(\frac{16}{20}\Big)^{25}\approx 0.40\%$$

This because, excluding the first and the last Yellow with probability $\Big(\frac{4}{20}\Big)^2$ the remaining $n=28$ draws can be represented by a binomial $Y\sim B(28;\frac{4}{20})$ and you have to calculate

$\Big(\frac{4}{20}\Big)^2\times \mathbb{P}[Y=3]$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.