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There is a bag of $10$ sweets with $5$ red, $3$ green and $2$ yellow. Dan chooses one at random and then keeps it, and then Carl chooses one.

What is the probability that Carl chooses a yellow sweet.

I don't know how to do this?

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  • $\begingroup$ Do you know about "conditional" probabilities? $\endgroup$
    – anak
    Jun 9, 2015 at 19:31

6 Answers 6

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You can do it using conditional probability, or if you don't know anything about it, then simply sum up the probabilities of the following disjoint events:

  • The probability that Dan chooses red and Carl chooses yellow is $\frac{5}{10}\cdot\frac{2}{9}=\frac{10}{90}$
  • The probability that Dan chooses green and Carl chooses yellow is $\frac{3}{10}\cdot\frac{2}{9}=\frac{6}{90}$
  • The probability that Dan chooses yellow and Carl chooses yellow is $\frac{2}{10}\cdot\frac{1}{9}=\frac{2}{90}$

So the probability that Carl chooses yellow is $\frac{10}{90}+\frac{6}{90}+\frac{2}{90}=\frac{18}{90}=\frac{1}{5}=20\%$.

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What's the probability that Carl chooses a yellow sweet given that Dan chose a sweet that isn't yellow?

What's the probability that Carl chooses a yellow sweet given that Dan chose a sweet that IS yellow?

That should be enough to help you do your homework.

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Hints: When Dan chooses a yellow sweet, the probability of Carl choosing one is $1/9$.

If Dan picks anything else, at that point, the probability of Carl choosing yellow is $2/9$.

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Either Dan chooses a yellow sweet or he doesn't, so we'll see what happens in each case.

If Dan chooses a yellow sweet, this is with probability $\frac{2}{10}$. So there are nine sweets left, one of which is yellow. What is the probability that Carl will choose a yellow sweet here? Multiply the two probabilities.

Now, if Dan chooses a non-yellow sweet, this is with probability $\frac{8}{10}$. So there are nine sweets left, two of which are yellow. What is the probability that Carl will choose a yellow sweet here? Multiply the two probabilities.

Now, you should have two numbers. Add them together.

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If Dan chooses a yellow one, then Carl has less chance, the chance of that, since $2$ of $10$ are yellow is $2/10=0,2$.

Now, if Carl chooses, in the first case, he needs to pick that $1$ yellow from the $9$ sweets, the chance of that is $1/9=0,111...$

In all other cases, Dan chooses something else, chance for that is $1-0,2=0,8$.

In the second case, he needs to pick $1$ from the $9$, but $2$ of them are yellow, so his chance is $2/9$.

If we sum these cases:: $0,2*0,111... + 0,8*0,222...$ Multiplying this with $100$ gives you the the answer in %

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The probability is $\frac2 {10}$. We know this because it doesnt matter what Dan chooses, he has an equal probability of eliminating anything, weighted based on their respective frequencies, i.e. \begin{align*} P(r)&=.5\\ P(g)&=.3\\ P(y)&=.2\\ \end{align*} so his choice does not effect the probability of anything more than anything else.

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  • $\begingroup$ Dan has an equal probability to eliminate red ($5$ out of $10$) and green ($3$ out of $10$)? $\endgroup$ Jun 9, 2015 at 19:45
  • $\begingroup$ Sorry about the imprecise statement of the reason, I am not sure how to state this in a clearer way, any suggestions would be much appreciated. $\endgroup$
    – nosyarg
    Jun 9, 2015 at 19:46
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    $\begingroup$ Put invisible distinct labels on the sweets. Any individual sweet is just as likely to be chosen first as to be chosen second. $\endgroup$ Jun 9, 2015 at 20:59

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