coin probability problem

sohailmbaprep wrote:Pls explain how to solve this:
q)the probability is 1/2 that a certain coin will turn up heads on any given toss.If the coin is to be tossed three times,what is the probability that on at least one of the tosses the coin will turn up tails ?

P(Atleast One Tail)
= 1 - P(No Tail, All Heads)
= 1 - (1/2)*(1/2)*(1/2)
=[spoiler]7/8[/spoiler]

Alternate/Long Method:
P(Atleast One Tail)
= P(1T,2H) OR P(2T,1H) OR P(3T)
= (3!/2!)/2*2*2 + (3!/2!)/2*2*2 + 1/2*2*2
= (3 + 3 + 1)/8
=[spoiler]7/8[/spoiler]

I know this doubt will arise, so let me try to pre-empt it.
Doubt: Why is there a 3!/2! on the numerator for P(1T,2H)?
Answer: The numerator is the Number of favourable outcomes for getting 1 Tail and 2 Heads. You can have 1 Tail and 2 Heads in the following cases:
T T H
T H T..
etc
The number of such ways = The number of ways of arranging T, T and H = (3!/2!)

aneesh.kg wrote:

sohailmbaprep wrote:Pls explain how to solve this:
q)the probability is 1/2 that a certain coin will turn up heads on any given toss.If the coin is to be tossed three times,what is the probability that on at least one of the tosses the coin will turn up tails ?

P(Atleast One Tail)
= 1 - P(No Tail, All Heads)
= 1 - (1/2)*(1/2)*(1/2)
=[spoiler]7/8[/spoiler]

Alternate/Long Method:
P(Atleast One Tail)
= P(1T,2H) OR P(2T,1H) OR P(3T)
= (3!/2!)/2*2*2 + (3!/2!)/2*2*2 + 1/2*2*2
= (3 + 3 + 1)/8
=[spoiler]7/8[/spoiler]

I know this doubt will arise, so let me try to pre-empt it.
Doubt: Why is there a 3!/2! on the numerator for P(1T,2H)?
Answer: The numerator is the Number of favourable outcomes for getting 1 Tail and 2 Heads. You can have 1 Tail and 2 Heads in the following cases:
T T H
T H T..
etc
The number of such ways = The number of ways of arranging T, T and H = (3!/2!)

Hi, aneesh,

Can we also use binomial probability distribution with this kind of problem?

If so, how?

Thank you,
Dave

I know that you've asked Aneesh but here's the answer. You definitely can. In fact, the way aneesh solved it in the second way is an application of the Binomial distribution.
For p=1/2=0.5 being the probability of a head, the probability of a tail is 1-p=0.5

Then the terms below show 3C0 term for no tails, 3C1 for one tail, and 3C3 for 3 tails.

In the case of 3 tosses, 1 = 3C0*0.5^0*0.5^3 + 3C1*0.5^1*0.5^2 + 3C2*0.5^2*0.5^1 + 3C3*0.5^3*0.5^0

So to find the answer, you could either subtract 3C0*0.5^3 from 1 or add the other 3 terms.

Let me know if this helps

eagleeye wrote:I know that you've asked Aneesh but here's the answer. You definitely can. In fact, the way aneesh solved it in the second way is an application of the Binomial distribution.
For p=1/2=0.5 being the probability of a head, the probability of a tail is 1-p=0.5

Then the terms below show 3C0 term for no tails, 3C1 for one tail, and 3C3 for 3 tails.

In the case of 3 tosses, 1 = 3C0*0.5^0*0.5^3 + 3C1*0.5^1*0.5^2 + 3C2*0.5^2*0.5^1 + 3C3*0.5^3*0.5^0

So to find the answer, you could either subtract 3C0*0.5^3 from 1 or add the other 3 terms.

Let me know if this helps

Thanks, eagleeye!

Sharp as ever!

eagleeye wrote:I know that you've asked Aneesh but here's the answer. You definitely can. In fact, the way aneesh solved it in the second way is an application of the Binomial distribution.
For p=1/2=0.5 being the probability of a head, the probability of a tail is 1-p=0.5

Then the terms below show 3C0 term for no tails, 3C1 for one tail, and 3C3 for 3 tails.

In the case of 3 tosses, 1 = 3C0*0.5^0*0.5^3 + 3C1*0.5^1*0.5^2 + 3C2*0.5^2*0.5^1 + 3C3*0.5^3*0.5^0

So to find the answer, you could either subtract 3C0*0.5^3 from 1 or add the other 3 terms.

Let me know if this helps

To add a little to what eagleeye said,
If there are 'n' number of tosses and 'r' number of these tosses must result in a Head, then
Required Probabilty = (n)C(r)/(2^n) = (n)C(n-r) /(2^n)

where,
nCr = (n)C(n-r) are the coefficients of Binomial Distribution.

But, having said all of that, 'Permutations & Combinations' becomes an interesting topic only when you understand the reason behind using nCr or (n!)/(r!)(n - r)! than just applying the above formula.

Both nCr and (n!)/(r!)(n - r)! are the number of ways of arranging r Heads and (n - r) Tails on the n tosses.

aneesh.kg wrote:

eagleeye wrote:I know that you've asked Aneesh but here's the answer. You definitely can. In fact, the way aneesh solved it in the second way is an application of the Binomial distribution.
For p=1/2=0.5 being the probability of a head, the probability of a tail is 1-p=0.5

Then the terms below show 3C0 term for no tails, 3C1 for one tail, and 3C3 for 3 tails.

In the case of 3 tosses, 1 = 3C0*0.5^0*0.5^3 + 3C1*0.5^1*0.5^2 + 3C2*0.5^2*0.5^1 + 3C3*0.5^3*0.5^0

So to find the answer, you could either subtract 3C0*0.5^3 from 1 or add the other 3 terms.

Let me know if this helps

To add a little to what eagleeye said,
If there are 'n' number of tosses and 'r' number of these tosses must result in a Head, then
Required Probabilty = (n)C(r)/(2^n) = (n)C(n-r) /(2^n)

where,
nCr = (n)C(n-r) are the coefficients of Binomial Distribution.

But, having said all of that, 'Permutations & Combinations' becomes an interesting topic only when you understand the reason behind using nCr or (n!)/(r!)(n - r)! than just applying the above formula.

Both nCr and (n!)/(r!)(n - r)! are the number of ways of arranging r Heads and (n - r) Tails on the n tosses.

Yes, you are right.

That being said, we could only use binomial probability distribution when we have just TWO events("success" and "failure", in this case: heads and tails), right?

But if we have three mutually exclusive events then the binomial probability can't be used. After all, the prefix "bi" means "two"

Ahhh.. this seems like a refresher on probability theory haha!