GMAT Math

Hi, there. I'm happy to put in my 2 cents here.

I agree with most of aneesh.kg's thoughts, but I agree with seal4913's calculations and numerical answers.

The very very rough rule of thumb is that, on probability, "and" means multiply and "or" means add. If that's where you understanding ends, you will quickly run into difficulties on the GMAT.

Here's a slightly more sophisticated treatment:

If events A & B are disjoint --- i.e. if they are mutually exclusive, with no possibility of happening together, then

P(A or B) = P(A) + P(B)

Two different faces of single die are disjoint. If I draw one playing card from a deck, two different suits would be disjoint.

If A and B are not disjoint, then you have to subtract the overlap which is counted twice when you add:

P(A or B) = P(A) + P(B) - P(A and B)

If events A & B are independent --- i.e. if the outcome of one has no impact on the outcome of the other, then:

P(A and B) = P(A)*P(B)

Two difference dice are independent, or one die and one spinner, or one die and the draw of one card. What color socks I wear and whether the stock market goes up that day are independent. Things that have absolutely nothing to do with one another are independent.

Of the questions above, #2 and #3 focus on very simple scenarios of independent events.
In #2, the coin and the spinner are independent.
In #3, the choice of the letter and the die are independent.
In both cases, you simply multiply the two probabilities, to get answers of 1/8 and 1/24 respectively, as seen in both posts above.

Question #1 is a bit trickier --- yes, the two coins are independent of one another, but now we have to consider multiple ways that a single result can arise. When I flip two coins, there are four possible outcomes --- if the the first entry is the result of the first coin and the second entry is the result of the second coin, then the outcomes are:
(H, H)
(H, T)
(T, H)
(T, T)

There's only one way to get double-heads or double-tails, but there are two ways to get 1 H, 1 T. Thus, the probability of getting 1 H, 1 T is 2/4 = 1/2.

In that example, we enumerated the four cases, but if you are flipping, say, six coins, and you want to know what the probability of getting 2H,4T, that will be too many combinations to enumerate. Here's how you would approach the more general problem.

Flip six coins (or flip one coin six times, which amounts to the same thing) --- for each flip, there are two possible outcomes, so there are 2^6 = 64 outcomes all together.

Of those, how many involve exactly 2 heads? Here, we have to use something called nCr (read "n choose r) --- the number 6C2 would tell us how many ways we could have six "slots" and choose exactly two of them (here, choosing those two "slots" to be heads). The way we calculate nCr is

nCr = [n!]/[(r!)((n-r)!)]

where n! (read "n factorial) means the product of all the positive integers from n down to one. Thus
2! = 2*1 = 2
4! = 4*3*2*1 = 24
6! = 6*5*4*3*2*1 = bigger than we need to think about without a calculator

6C2 = [6!]/[(2!)*(4!)] = [6*5*4*3*2*1]/[(2)*(4*3*2*1)]= 6*5/2 = 15

Incidentally, that massive cancellation always happens in computing nCr. Also, if it means anything to you, the nCr numbers are also the numbers on Pascal's Triangle.

Thus, the probability of flipping six coins and getting exactly 2 heads in the six tosses is 15/64.

Does all this make sense? Here's another probability question for practice:
https://gmat.magoosh.com/questions/839
When you submit your answer, the next page will have a video explanation of the question.

Let me know if you have any further questions.

Mike