Math Question?

[swellguy]

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?

---

Members don't see this ad.

 

[thebozz1975]

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]

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]

FYI, vvvv's explanation is wrong. See mine above.

 

[t123456]

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

 

[thebozz1975]

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]

is there anyy way of actually calculating the answer instead of sitting there writing out all the possibilities?

 

[Streetwolf]

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

 

[thebozz1975]

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]

(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]

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]

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]

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.