Hi infiniti007,

In probability questions, there are two results that you can calculate - what you WANT to have happen or what you DON'T want to have happen. Since there are so many different ways to flip 2, 3, 4 or 5 tails, it will be easier for us to calculate what we DON'T want (0, 1 or 6 tails).

Since each toss has 2 possible outcomes (heads or tails), there are 2^6 = 64 different results for 6 coin flips.

Of those 64 options...

0 tails -->
HHHHHH = 1 option

1 tail -->
THHHHH
HTHHHH
HHTHHH
HHHTHH
HHHHTH
HHHHHT = 6 options

6 tails -->
TTTTTT = 1 option

1 + 6 + 1 = 8 options (of the 64) that we DON'T want...

Thus 64/64 - 8/64 = 56/64 = 7/8 that we DO want.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

Here it's best to calculate the probability that 0, 1, or 6 heads appear and subtract that from 1.

One of the benefits of this approach is that you only have to calculate the probability of 3 results rather than 4 results (2, 3, 4, or 5 heads). But the real payoff is the simplicity of the calculations.

No heads can only happen on one way. One head can happen in 6C1 = 6 ways. Finally, all heads (no tails) can only happen in one way. So, 0, 1, or 6 heads can happen in 1 + 6 + 1 = 8 ways. Each individual outcome has probability (1/2)^6 = 1/64. So these 8 particular outcomes have a collective probability of 8/64 = 1/8. Finally we subtract this from the sum of all probabilities to get 7/8. (C)

This is an example of a Bernoulli trial. A Bernoulli trial only has two results - success and failure (like heads or tails, or rolling a die and getting a 6 or rolling a die and not getting a 6). If each of N trials is an independent event and the probability of a success is P, then the probability of R successes in N trials is

(NCR)(P)^R(1-P)^(N-R)

(NCR) tells us the number of ways we can have R successes in the N trials. Because all the trials are independent we know that R successes results in R factors of P, and N-R factors of 1-P. We get all the possibilities in one package.

For example if John rolls a six sided fair die 5 times what is the probability that an odd prime will be rolled 2 or 3 times.

We call rolling an odd prime (3 or 5) a success, and rolling anything else a failure. So the probability of success is 2/6 = 1/3. Consequently, the probability of failure is 2/3

Three successes: (5C3)(1/3)^3(2/3)^2 = (5!)/(3!2!)(1/27)(4/9) = ((5*4)/2)(4/243) = 40/243
Two successes: (5C2)(1/3)^2(2/3)^3 = (5!)/(3!2!)(1/9)(8/27) = ((5*4)/2)(8/243) = 80/243
Two OR three successes: (40 + 80)/243 = 120/243

Bernoulli trials come up in many probability questions and it's useful to be familiar with the concept so you can avoid listing all the iterations of certain outcomes. In the original problem listing possibilities is a reasonable approach, but a small changes (more trials, unevenly weighted outcomes, etc.) would make it impracticable.