I know its simple, but this is melting my face... What is the probability of getting at most one head in three coin tosses? I used the equation nCk (P^k)(q^n-k) --- I keep getting 7/8 but this book says the answer is 1/2, what am I screwing up?

I know its simple, but this is melting my face... What is the probability of getting at most one head in three coin tosses? I used the equation nCk (P^k)(q^n-k) --- I keep getting 7/8 but this book says the answer is 1/2, what am I screwing up?

I dont think you should sit there and count the pr. at most = 1 head and 0 head.

n = number of trials r = number of specific events you wish to
obtain p = probability that the event will occur q = probability that the event will not occur
(q = 1 - p, the complement of the event

Use this formular to cal pr of 1 and 0 then plus them together. In this case you can cal at least, at most with any number...

Hopefully those combinations would be intuitive and you could answer that question very quickly without any numerical computation. I only wrote them out for the benefit of those who don't "see it".

First calculate the probability of gettng 0 heads which is 1/8 then add it to the probability of getting 1 head which is 3c1*.5^1*.5^2 which equals 3/8. (--->the success failure formula). Add the two together o get 4/8. Its very simple.

First calculate the probability of gettng 0 heads which is 1/8 then add it to the probability of getting 1 head which is 3c1*.5^1*.5^2 which equals 3/8. (--->the success failure formula). Add the two together o get 4/8. Its very simple.

So then calculate the proability of getting 3 and then subtract it from 1. Therefore you would do 1/2*1/2*1/2=1/8 Now 1-(1/8)=(7/8). The answer is therefore 7/8.