math coin problem

[Electrons]

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A coin is tossed 5 times. What is the probability that you get 1 head AND 4 tails?

Answer: 5/32.



I guessed it correctly, but no idea how it should be setup or how to reason it.

 

[PooyaH]

A coin is tossed 5 times. What is the probability that you get 1 head AND 4 tails?

Answer: 5/32.



I guessed it correctly, but no idea how it should be setup or how to reason it.

These are how you'd get 1H and 4Ts:

H-T-T-T-T
T-H-T-T-T
T-T-H-T-T
T-T-T-H-T
T-T-T-T-H

All possible outcomes = P^n...n = number of consecutive events (in this case 5)...P = Number of possible outcomes (in this case 2 since you can only get H or T)

2^5 = 32

So

Probability = 5/32

 

[klutzy1987]

These are how you'd get 1H and 4Ts:

H-T-T-T-T
T-H-T-T-T
T-T-H-T-T
T-T-T-H-T
T-T-T-T-H

All possible outcomes = P^n...n = number of consecutive events (in this case 5)...P = Number of possible outcomes (in this case 2 since you can only get H or T)

2^5 = 32

So

Probability = 5/32

That is way too long to do on the DAT. You have to use the Success/Failure formula.

NcR * (Prob of sucess)^#of succeses * (prob failure)^#failure

If you post an exact question i will demonstrate how this formula is used.

 

[Orgodox]

That is way too long to do on the DAT. You have to use the Success/Failure formula.

NcR * (Prob of sucess)^#of succeses * (prob failure)^#failure

If you post an exact question i will demonstrate how this formula is used.

Yea I agree, its fine to do other things here but sometimes its complicated and this formular works. thanks for showing me how to use it btw.

Here it would be
5!/1!4! (1/2)^4 (1/2)^1
which is 5 (1/2)^5= 5 (1/32)= 5/32

 

[klutzy1987]

Yea I agree, its fine to do other things here but sometimes its complicated and this formular works. thanks for showing me how to use it btw.

Here it would be
5!/1!4! (1/2)^4 (1/2)^1
which is 5 (1/2)^5= 5 (1/32)= 5/32

Bingo. I forgot the question was up there lol. Perfecly done though, i see that youve got that one down.

 

[Orgodox]

Im working on it. Doing TONS of math caze I need the practice. Got my time down after a few practice math tests and now just have to improve on accuracy. I dont think the trig is worth it everyone says its hard even i you know it so Ill just guess on my few I get and that way ill have the few extra mins which will allow me to have the time to ensure I can get the ones I know right.
Can you apply this formula to the coin problem like 4 coins flipped whats the chances of getting 2 heads in a row?

 

[klutzy1987]

No for such a problem you would need to calculate the probability of getting 2 heads in a row on the first 2 spins and then multiply by however many possible ways there are of getting it. I think the answer to that would be (1/2)(1/2)*3

 

[Electrons]

Thanks guys.

No for such a problem you would need to calculate the probability of getting 2 heads in a row on the first 2 spins and then multiply by however many possible ways there are of getting it. I think the answer to that would be (1/2)(1/2)*3

Klutzy, can you explain the difference. They look same problem to me. Mine is like 1 head then 4 tails in a row. His is just 2 heads in a row. How come you have to do something different to it for his?

 

[klutzy1987]

The difference is that it asked for 2 in a row. Therefore it eliminates alot of possibilities. If the question would have just asked for 2 heads and 3 tails then yu would have used the formula i posted bfore.