# Probability Question: Mutually Exclusive Events

Source of original question and answers can be found here under "Exercise 1" http://www.intmath.com/counting-probability/9-mutually-exclusive-events.php

A box contains 100 items of which 4 are defective. Two items are chosen at random from the box. What is the probability of selecting:

(a) 2 defectives if the first item is not replaced;

(c) 1 defective and 1 non-defective if the first item is not replaced?

I am having issues understanding (c). When I did (a), it was a simple dependent events kind of problem, initially there are 4 defectives, you pick one out, and now there are 3 defectives out of 99.

However, when I did (c), I reasoned through it similar to how I reasoned through (a), and my math was:

(4/100) X (96/99).

This was wrong as the answer said we need to account for the other possibility: what if a non-defective was picked first and then a defective. (C) reads to me very similar to (A). The difference is there is an "and", someone told me it is commutative, doesn't matter what comes first, you must account for all possibilities.

But I get my head confused about other "and" probability questions. This one categorized as a mutually exclusive example. But when I think of other AND dependent probability questions I've done, I didn't have to solve for the other possibility. For example: colored marble problems, there are X blue marbles and Y red marbles in a bag, what are the chances of pulling a blue and then a red without replacement. Those questions only require to consider the first scenario, not account for "oh what if I picked a red marble first instead of blue?"

When I read problem (c) it seems that order does matter, the first is defective, not maybe it could be the other way.

Am I reading it wrong, if the problem instead said "the first item as defective and the second as non-defective if the first item is not replaced" would my original answer be correct?

• You can think of it as in this problem we are choosing them simultaneously. While, yes, you can arrive at the correct answers by temporarily assuming order matters, we could also approach this the other way. By making a temporary assumption that all items are labeled, there are $\binom{100}{2}$ different equiprobable outcomes for choosing two items simultaneously, of which, $\binom{4}{2}$ will have both defective and $\binom{4}{1}\binom{96}{1}$ have exactly one defective. $\binom{n}{r}=\frac{n!}{r!(n-r)!}$ is the binomial coefficient Jul 21, 2016 at 19:45
• @JMoravitz I'm new to this site, I hope this reply reaches you. So okay, if we think of it as simultaneous drawings of the item, yes, I can agree with their explanation of the problem. Regarding, the second part of your reply "temporarily assuming order matters", I'm not following you. I am studying for the GRE, my knowledge of probability is high school level so the n, r and notation you are using is making it a bit too beyond me to understand. Is it possible you can re-explain using fractions?
– Acer
Jul 21, 2016 at 21:28
• That is precisely why I linked to the binomial coefficient wikipedia page. The point I'm trying to make is that it is more intuitive to use the binomial coefficients for this problem than it is to use direct counting methods. $\binom{n}{r}$ is the number of ways to take $r$ items out of $n$ total. $\binom{100}{2}$ is the number of ways of taking two things out of 100 total. By temporarily assigning labels to the items, we have $\binom{100}{2}=495$ equally likely outcomes (as opposed to three outcomes which are not equiprobable). Jul 21, 2016 at 21:50
• Typo, I'm sorry, $\binom{100}{2}=4950$. That's what I get for trying to calculate it in my head. One final comment and then I'm out to cook dinner. The answer I proposed was $\binom{4}{2}/\binom{100}{2} = 6/4950=\frac{1}{825}=\frac{4}{100}\cdot \frac{3}{99}$, so the answer is the same. Similarly, for part c) $\binom{4}{1}\binom{96}{1}/\binom{100}{2}=\frac{4\cdot 96}{4950}=\frac{4}{100}\cdot \frac{96}{99}+\frac{96}{100}\cdot \frac{4}{99}$, so again, the answers are the same Jul 21, 2016 at 22:00
• Binomial coefficients are more useful in my opinion whenever we are referring to sets or other objects in which order of selection doesn't matter. Most of those questions you can temporarily assume order does matter and still get an answer correctly, but a great deal of work can be saved by ignoring how things are organized. For example, calculating the odds of a full house in a five-card poker hand is $\binom{13}{1}\binom{12}{1}\binom{4}{3}\binom{4}{2}/\binom{52}{5}$ Jul 21, 2016 at 22:03