Eli has 7 black pens and 5 blue pens in his desk drawer. he also has 3 yellow highlighters, 2 green highlighters, and 5 pink highlighters in his pencil case. If he chooses one pen and one highlighter without looking, what is the probability that he will get a black pen and he will not get a green highlighter?

A. 7/15
B. 117/1,000
C. 1/3
D. 1/12

3 answers

7/12 * 8/10
For this problem, you have to first separate the desk drawer and the pencil case.

If Eli picked out the desk drawer, there is 7/12 chance that he gets a black pen. The 7 comes from the number of black pens, the 12 comes from the total number of pens. The reason why we make the fraction the way it is that you assume that all pens have a equal chance of getting picked, so because 7/12 of the pens are black, it is like saying if I picked 12 pens, 7 of them would be black, so that fraction tells that same thing if I picked one pen.

You multiply this probability by the probability you will not get a green highlighter (which is the same probability as picking a pink highlighter). You multiply them because first Eli has to get a black pen, then after that, Eli has to pick a pink highlighter, so it is like you take the first probability and you want the second probability to be a certain fraction 'of' the first one. The 'of' is like 1/2 of a 1/2. You take 1/2 and you want to find a half of that. 1/2 times 1/2 = 1/4. In this case, instead 1/2 of 1/2, its 7/12 times the probability of picking pink.

I think you have it from here :) Just remember to find the total number of highlighters. The probability will be a part / whole.
7/15