# Math Question?

#### swellguy

10+ Year Member
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?

#### thebozz1975

10+ Year Member
5+ Year Member
These are the possible outcomes:

HHT
HTH
HHH
TTT
THH
HTT
THT
TTH

4/8 or 1/2 of those outcomes have zero or one head.

#### vvvv

10+ Year Member
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...

#### thebozz1975

10+ Year Member
5+ Year Member
FYI, vvvv's explanation is wrong. See mine above.

#### t123456

10+ Year Member
getting at most 1 head, most is the key word, so there are only 3 possibilities

#### thebozz1975

10+ Year Member
5+ Year Member
t1234's explanation is wrong. There are 4, not 3 possibilities. You could get zero heads, or a head on the first, second, or third throw.

4 possibilities not 3. Just see my explanation above.

#### userah

10+ Year Member
7+ Year Member
is there anyy way of actually calculating the answer instead of sitting there writing out all the possibilities?

#### Streetwolf

##### Ultra Senior Member
10+ Year Member
7+ Year Member
(3c0)(1/2)^0(1/2)^3 + (3c1)(1/2)^1(1/2)^2 = 1/8 + 3/8 = 1/2.

#### thebozz1975

10+ Year Member
5+ Year Member
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".

#### Electrons

10+ Year Member
(3c0)(1/2)^0(1/2)^3 + (3c1)(1/2)^1(1/2)^2 = 1/8 + 3/8 = 1/2.
If it asked at most two heads, then it would be this right?

(3c0)(1/2)^0(1/2)^3 + (3c1)(1/2)^1(1/2)^2 + (3c2)(1/2)^2(1/2)^1= 1/8 + 3/8 + 3/8= 7/8

#### klutzy1987

##### StudyingSucks Letsgo Mets
10+ Year Member
5+ Year Member
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.

#### Electrons

10+ Year Member
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.
Klutzy, I get that from the formula. I was wondering what the answer is if it asked "at most 2" instead of "at most 1".

#### klutzy1987

##### StudyingSucks Letsgo Mets
10+ Year Member
5+ Year Member
Klutzy, I get that from the formula. I was wondering what the answer is if it asked "at most 2" instead of "at most 1".
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.