Probability PS

gmatstud25 wrote:what is the probability that a coin lands tail on exactly the 4 attempt in a total of 6 attempts?

The probability of getting exactly k results out of n flips is:
nCk/2^n

answer 15/64?

4meonly wrote:

gmatstud25 wrote:what is the probability that a coin lands tail on exactly the 4 attempt in a total of 6 attempts?

The probability of getting exactly k results out of n flips is:
nCk/2^n

answer 15/64?

Perfect!

To elaborate a bit:

6C4/2^6 = (6!/4!2!)/64 = (6*5*4*3*2/4*3*2*2)/64 = 15/64

The question is "what is the probability of getting tails on the 4th attempt and not getting exactly 4 tails in 6 attempts"

I think the probability should still be 1/2.

Hi Stuart,
At your earliest convenience could you please look in to the coin flip strategy post and let us know how to do it.

We are not sure on how to use the pascals triangle approach there. Its one post above this one.

Thanks!

gmatstud25 wrote:The question is "what is the probability of getting tails on the 4th attempt and not getting exactly 4 tails in 6 attempts"

I just assumed that the question had been misquoted.

If the question is actually "what is the probability of getting tails on the 4th attempt out of 6 attempts", then the answer would be 1/2, since the probability of getting tails on any flip is 1/2, assuming it's a "fair coin".

I'm pretty sure the question should have been "If a fair coin is flipped 6 times, what's the probability of getting tails on exactly 4 of the 6 flips".

cramya wrote:Hi Stuart,
At your earliest convenience could you please look in to the coin flip strategy post and let us know how to do it.

We are not sure on how to use the pascals triangle approach there. Its one post above this one.

Thanks!

I'm not sure which post you mean, sorry.

I should have been more specific.

This is the post I am referring to
https://www.beatthegmat.com/coin-flip-st ... 19517.html

Thanks!

The coin flip shortcut ( which uses Combinations ) is a specific case of a Generalization called Binomial theorem.

BT says : If probability of success of event is X and probability of failure is Y and if we want N successes out of M tries then the resultant probability is

mCn * (x)^n * (y)^(m-n)