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There are $4$ eyeliners and $5$ lip liners in your makeup bag. If you select a product at random from your bag, what is the probability that you will select one lip liner and then one eyeliner out from your makeup bag?

This is a fairly simple question but I am a little confused about which approach is correct should my solution be:

$\frac{4}{9} \times\frac {5}{8}$ because I have $5$ options of getting an lip liner and then $4$ options of getting a lip liner for a total of $\frac{20}{72}$

or should the answer be $\frac{1}{5} \times \frac{1}{4}$ because I only want one lip liner out of the $5$ and only one eyeliner out of the $4$ and then $\frac{1}{20}$

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    Your first answer is correct. I don't understand the second...even if you specified which eyeliner and which lip liner you wanted, that wouldn't be the answer. – 2017-01-19
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    Your 2nd suggested answer would be right if 5 different lip liners were in your lip liner bag, and 4 different eye liners were in your eye liner bag, and you wanted to find the probability that you first chose exactly the right eye liner and then chose exactly the right lip liner. – 2017-01-19

2 Answers 2

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For success, you must choose lip liner first, and then eye liner 2nd. The probability of success is the product of being successful on both occasions.

Probability of success on the first occasion is $\frac{4}{9}$

Probability of success on the 2nd occasion is $\frac{5}{8}$

The product of these is $\frac{20}{72}=\frac{5}{18}$

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Your first answer is correct.

I have tried to explain the problem in a little detail as below:

This is a very simple problem involving combinations.

Let A be the event of drawing 1 lip liner, Let B be the event of drawing 1 eye liner

Clearly, since the objects are not replaced before the second draw, then the events A and B are not independent and so, the required probability is:

P(A∩B)= P(A)P(B|A) = (5/9)(4/8) = 20/72

where P(A) = $\binom{5}{1}$ / $\binom{9}{1}$

and P(B|A) = $\binom{4}{1}$ / $\binom{8}{1}$

where P(A) is such because, in the first draw, out of the total 9 choices (=5 lip liners + 4 eye liners) I have 5 choices when drawing a lip liner. However, since the experiment is without replacement, my total number of choices after the first draw reduce to 8 (out of which I have 4 lip liner and 4 eye liners). So, now the probability of drawing one eye liner given I have already drawn one lip liner without replacement (denoted by P(A|B)) will be 4 choices ( because there are 4 eye liners) out of the total 8 available.