1
$\begingroup$

Each of the eight customers are choosing randomly and independently on each other one flavour of candy. In the bowl there many candies with five different flavours.

a) What is the probability that each flavour is chosen? And what would happen, if there were 100 customers?

b) What is the probability that all of the customers has agreed on the same flavour?

a) $P=1 - (\frac{1}{5})^{8} + (\frac{2}{5})^{8} + (\frac{3}{5})^{8} + (\frac{4}{5})^{8}$

My intuition is that one wasn't chosen , two wasn't chosen, three wasn't chosen ...

Is it correct?

For 100 customers it is the same just exponent is equal to 100.

b) It is just $P=(\frac{1}{5})^8$ right?

  • 0
    b. is wrong....there are $5$ possible colors they might have agreed on so you need to multiply by $5$. Likewise in a. you need to keep track of the number of choices for the omitted colors. – 2017-01-22
  • 0
    Note: nothing in the problem says that the flavors are equiprobable, that's just something you guessed at. Granted, you have to assume something and this is probably the best available guess...but it's just a guess. That should have been specified in the problem but, failing that, you should clearly indicate that you have added that assumption. – 2017-01-22

1 Answers 1

0

I think in the first case it should be $$P=1-\binom 51(\frac{4}{5})^8+\binom 52(\frac{3}{5})^8-\binom 53(\frac{2}{5})^8+\binom 54(\frac{1}{5})^8$$

Here I have utilised the Inclusion Exclusion Principle where the property is taken as a particular flavour is not chosen.

b) is incorrect.

Edit after @lulu comment to b)below the correct answer should be $5(\frac{1}{5})^8$.Read @lulu comment below for explanation.

  • 1
    Like the OP, you are neglecting to keep track of the number of ways to make the appropriate omission. For example, in part $b$, there are five possible colors the team might have coincided on, so the answer should be $5\times \left(\frac 15\right)^8$. And so on. – 2017-01-22
  • 0
    @lulu we are finding the probability not the number of ways.All people can agree on 1 flavour of all 5 flavours.So there should be no problem with b) – 2017-01-22
  • 0
    I understand it's a probability (my answer is obviously a probability, not a count). They can agree on White, Red, Blue, Green, or Yellow. Each of those has probability $\left( \frac 15 \right)^8$ and there are five cases, so you must multiply by $5$. – 2017-01-22
  • 0
    @lulu what about a).Do you have anything to say about it. – 2017-01-22
  • 0
    Yes, it's wrong for the same reason. You need to count the ways the various states can occur. For example the number of ways to miss at least two colors is $\binom 52 \times \left( \frac 35 \right)^8$. – 2017-01-22
  • 0
    Think of it this way: if you and each toss a fair coin the probability that we get the same result is $\frac 12$ not $\frac 14$. That's because we might get $HH$ or $TT$ so the answer is $2\times \left( \frac 12 \right)^2$. – 2017-01-22
  • 0
    @lulu suppose instead of eight persons we have two persons.The probability according to you is $(1/5)^2*5$ which is $1/5$.you can yourself see that it is not possible. – 2017-01-22
  • 0
    Of course the answer for two people would be $\frac 15$. The first person chooses whatever and the second matches that choice with probability $\frac 15$. – 2017-01-22
  • 0
    If you need to enumerate, then, using my colors, the cases in which two people match are $WW, RR,BB,GG,YY$. Each of these has probability $\frac 1{25}$ so the answer is $5 \times \frac 1{25}=\frac 15$. – 2017-01-22
  • 0
    @lulu Thanks a lot.I was mistaken.Please do post your comments as an answer. – 2017-01-22
  • 0
    I prefer to leave the work to the OP...the question looks like a homework problem to me so I don't want to do more than leave hints. If you feel differently, please go ahead and modify your posted solution. Worth noting: nothing in the problem says that the colors are equiprobable. Granted, there's not much to say if you don't assume something like that, but that's a very strong assumption to simply impose. It should really be clarified by whomever set the problem but, failing that, the assumption ought to be spelled out. – 2017-01-22