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40% of students have brown eyes what are the chances if at random 2 are chosen that at least 1 has brown eyes. The answer is 64% and I don't know why?!

Jul 19th, 2014

Let's go ahead and list every possible situation. There are four of them, where eye colors of student 1 and student 2 are listed next to each other:

Brown , Brown

Brown , Not Brown

Not Brown , Brown

Not Brown , Not brown

Notice that the second and third outcomes are the same - the order of which student has the Not Brown eye color doesn't matter. Since the problem is independent of the order you write Student 1 and Student 2, we need to  look at combinations instead of permutations. So, we can take the probability of the second and third option and just multiply by 2 because their probabilities are the same anyway:

Brown , Brown

Brown , Not Brown  OR reverse order, x2 because there are 2 ways to rearrange it without changing probability

Not Brown , Not brown

Now, let's calculate the probability. Just multiply both probabilities in each possibility listed. Why multiplication? This is because some probabilities can be shown in a diagram as a tree strucutre, with the first event as the root and subsequent events as branches which branch out into so called "leaves" of final probabilities. Every time you have a possibility that leads to other possibilities, you have a tree structure and you multiply the probabilities. If the question asked what the probability was that a student would have brown eyes and brown hair, you would add the two probabilities instead of multiplying, because one of the events has no influence on the other. Here are the calculations:

Brown, Brown prob. = 40% * 40% or 16%

Brown, Not brown = 40 * (100% - 40%) or 24%, and times 2 so it is 48%

Not Brown, Not Brown = (100% - 40%) * (100% - 40%) or 36%

Let's check that these probabilities add up to 100%, since they have to because we have listed every possible outcome here and probabilities always add up to 100%. So, 16% + 48% + 36% = 100%

Now let's select the probability we are interested in. We want to know the chance of at least one brown eyed person. We can either add up 16% and 48%, since these two outcomes (non-branching at this point) are independent, or we can do 100% - 36% since we are interested in every probability EXCEPT that one and they all add up to 100%. Either way (16%+ 48% or also 100% - 36%) you will get the same answer of 64%.

Jul 19th, 2014

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Jul 19th, 2014
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Jul 19th, 2014
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