Probability of Getting 4 Tails in 10 Flips

[Miike012]

You flip a count 10 times what is the probability of getting exactly four tails?

My solution.
out of ten flips you can get

0 tails out of 10 flips
1 tails out of 10 flips
2 tails out of ten flips
3 tails out of ten flips
4 tails out of ten flips
5 tails out of ten flips
6 tails out of ten flips
7 tails out of ten flips
8 tails out of ten flips
9 tails out of ten flips
or
10 tails out of ten flips

So there are 11 different cases. Therefore would the answer be 1/11?

 

[Dick]

You flip a count 10 times what is the probability of getting exactly four tails?

My solution.
out of ten flips you can get

0 tails out of 10 flips
1 tails out of 10 flips
2 tails out of ten flips
3 tails out of ten flips
4 tails out of ten flips
5 tails out of ten flips
6 tails out of ten flips
7 tails out of ten flips
8 tails out of ten flips
9 tails out of ten flips
or
10 tails out of ten flips

So there are 11 different cases. Therefore would the answer be 1/11?

Do you really think those cases are equally probable??

 

[phinds]

As Dick suggests, you need to rethink the problem. What is the probability of NOT getting a tail on any single flip?

 

[Ray Vickson]

You flip a count 10 times what is the probability of getting exactly four tails?

My solution.
out of ten flips you can get

0 tails out of 10 flips
1 tails out of 10 flips
2 tails out of ten flips
3 tails out of ten flips
4 tails out of ten flips
5 tails out of ten flips
6 tails out of ten flips
7 tails out of ten flips
8 tails out of ten flips
9 tails out of ten flips
or
10 tails out of ten flips

So there are 11 different cases. Therefore would the answer be 1/11?

You are missing some basics, so to clarify, start with the simpler case of 2 flips. You can write down all the possible ways to obtain 0 tails, 1 tail and 2 tails, and from there you can get the probabilities.

You need to assume something about the flips: (i) they are fair (equal chance of H or T each time); and (ii) the flips are independent (that is, the results on one flip do not affect the probabilities of H or T on any other flip---the coin does not "remember" anything).

 

[Miike012]

You are missing some basics, so to clarify, start with the simpler case of 2 flips. You can write down all the possible ways to obtain 0 tails, 1 tail and 2 tails, and from there you can get the probabilities.

You need to assume something about the flips: (i) they are fair (equal chance of H or T each time); and (ii) the flips are independent (that is, the results on one flip do not affect the probabilities of H or T on any other flip---the coin does not "remember" anything).

0 tails:
Flip 1: H
Flip 2: H

1 tails:
Flip 1: H
Flip 2: T

OR

Flip 1: T
Flip 2: H

2 tails:
Flip 1: T
Flip 2: T

There are 4 cases.. is that correct?

 

[Dick]

0 tails:
Flip 1: H
Flip 2: H

1 tails:
Flip 1: H
Flip 2: T

OR

Flip 1: T
Flip 2: H

2 tails:
Flip 1: T
Flip 2: T

There are 4 cases.. is that correct?

Yes, so for two coins that shows the probability of 0 tails is 1/4, 1 tail is 2/4 and 2 tails is 1/4. It's going to be tedious to write out cases that way with 10 coins. Haven't you learned to use combinatorial arguments or the binomial distribution to calculate it without writing cases out?

 

[Miike012]

Yes, so for two coins that shows the probability of 0 tails is 1/4, 1 tail is 2/4 and 2 tails is 1/4. It's going to be tedious to write out cases that way with 10 coins. Haven't you learned to use combinatorial arguments or the binomial distribution to calculate it without writing cases out?

So then it would be

10 choose 0
10 choose 1
10 choose 2
.
.
.
10 choose 10
Then would the answer be

10 choose 4/(Ʃ(10 choose i)) i from 0 to 10

 

[Dick]

So then it would be

10 choose 0
10 choose 1
10 choose 2
.
.
.
10 choose 10
Then would the answer be

10 choose 4/(Ʃ(10 choose i)) i from 0 to 10

Yes, it would. But there's an easy way to evaluate the sum in the denominator. It's all total ways of choosing heads or tails. It's 2^10.

 

[Miike012]

Yes, it would. But there's an easy way to evaluate the sum in the denominator. It's all total ways of choosing heads or tails. It's 2^10.

Thank you!

 

[<sHoRtFuSe>]

Let p=Probability of getting Heads=0.5
q=Probability of getting Tails=0.5
assuming all flips are independent, probability of getting 4 heads=p*p*p*p*q*q*q*q*q*q
you are doing the mistake of assuming getting 1, 2, 3 ...heads(or tails) is equally likely. They clearly aren't. This is a typical example of bernoulli trials.

 

[haruspex]

Let p=Probability of getting Heads=0.5
q=Probability of getting Tails=0.5
assuming all flips are independent, probability of getting 4 heads=p*p*p*p*q*q*q*q*q*q

Well, no, that's the probability of getting four heads first then six tails. And it's the same as any other specific sequence of outcomes.

 

[<sHoRtFuSe>]

Well, no, that's the probability of getting four heads first then six tails. And it's the same as any other specific sequence of outcomes.

Okay i should have multiplied by 10!/4!6!

 

[PWiz]

You flip a count 10 times what is the probability of getting exactly four tails?

My solution.
out of ten flips you can get

0 tails out of 10 flips
1 tails out of 10 flips
2 tails out of ten flips
3 tails out of ten flips
4 tails out of ten flips
5 tails out of ten flips
6 tails out of ten flips
7 tails out of ten flips
8 tails out of ten flips
9 tails out of ten flips
or
10 tails out of ten flips

So there are 11 different cases. Therefore would the answer be 1/11?

You also have to consider the order in which you get the heads and tails, and this problem would clearly become tedious without the binomial distribution, or without applying permutations. What I mean is that you'll have to consider the probability of ALL the different ways tails in the combination T,T,T,T,H,H,H,H,H,H could be arranged. (Like having H,T,T,H,T,TH,H,H,H is different and its probability also has to be included in the final answer.)