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Question: A shipment of milk chocolate consists of 500 chocolate bars. The buyer tries three randomly selected bars, and accepts the shipment if each of the three is indeed a milk chocolate. Find the probability that the shipment is accepted if it has 498 milk and 2 dark chocolate bars.

Attempt:

I know that the overall selection will be C(500,3).

Probability of selecting a dark chocolate bar is 1/250 --> fails

Probability of selecting a milk chocolate bar is 249/250 --> accepted

Help:

How do I put these elements together to find accepted probability? Do I multiple C(500,3) by the accepted percentage of 249/250?

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    $\begingroup$ There are $498$ milk chocolate bars and $2$ dark chocolate bars. It is accepted if all three bars selected are milk chocolate. You have four-hundred and ninety-eight milk chocolate bars and you want to choose three of them. How many ways can this be done? $\endgroup$ – JMoravitz Feb 25 '18 at 18:35
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    $\begingroup$ As an aside, you should have immediately known that $\binom{500}{3}\times \frac{249}{250}$ could not have been correct since this number is larger than $1$ and could not have been a probability. $\endgroup$ – JMoravitz Feb 25 '18 at 18:36
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If you are using a counting argument then indeed there are ${500 \choose 3}$ equally likely ways of choosing $3$ from $500$ bars. Of these, there are ${498 \choose 3}$ ways of choosing $3$ from $498$ milk chocolate bars, so you can just divide one number by the other

Alternatively, the probability the first is milk chocolate is $\frac{498}{500}$. Given that, the conditional probability the second is also milk chocolate is $\frac{497}{499}$. Given those two both being milk chocolate, the conditional probability the third is also milk chocolate is $\frac{496}{498}$. Multiply these together

Either way, you will get a probability of $\frac{498 \times 497 \times 496}{500 \times 499\times 498}\approx 0.988$ the shipment is accepted

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