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Head-Tail Question

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by goelmohit2002 » Thu Aug 06, 2009 9:51 pm
Hi All,

Probability of getting a head on a toss of non fair coin is 0.6. In 6 tosses of a coin, what is the probability of getting atleast 5 heads ?

OA = Will post after some discussion.

Thanks
Mohit
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by real2008 » Thu Aug 06, 2009 10:00 pm
case1) five heads and six ways (0.6^5*0.4^1*6)

case2) six heads one way=(0.6^6)

probability= case1+ case2 =0.23328

OA pl?
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by goelmohit2002 » Thu Aug 06, 2009 10:03 pm
real2008 wrote:case1) five heads and six ways (0.6^5*0.4^1*6)

case2) six heads one way=(0.6^6)

probability= case1+ case2 =0.23328

OA pl?
Please think more :-)

OA is different .....Actually I also was puzzled how to reach to that answer....
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Re: Head-Tail Question

by ankitns » Thu Aug 06, 2009 10:31 pm
Case 1 - 5 heads:

(6/10)^5 * (4/10) = (6^5 * 4)/(10^6)


Case 2 - 6 heads = (6^6)/(10^6)

Adding the two cases gives us (6^5)/(10^5)

which can be simplified to (3/5)^5...or (0.6)^5


Whats the OA?

Thanks.

goelmohit2002 wrote:Hi All,

Probability of getting a head on a toss of non fair coin is 0.6. In 6 tosses of a coin, what is the probability of getting atleast 5 heads ?

OA = Will post after some discussion.

Thanks
Mohit
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
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by goelmohit2002 » Thu Aug 06, 2009 10:36 pm
goelmohit2002 wrote:
real2008 wrote:case1) five heads and six ways (0.6^5*0.4^1*6)

case2) six heads one way=(0.6^6)

probability= case1+ case2 =0.23328

OA pl?
Please think more :-)

OA is different .....Actually I also was puzzled how to reach to that answer....
Sorry Real2008...I missed your 6 in case 1.....

You are indeed correct....can you please tell why are you multiplying the same by 6...why not the solution be as given by "ankitns"...above ?
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by tohellandback » Thu Aug 06, 2009 10:40 pm
goelmohit2002 wrote:
goelmohit2002 wrote:
real2008 wrote:case1) five heads and six ways (0.6^5*0.4^1*6)

case2) six heads one way=(0.6^6)

probability= case1+ case2 =0.23328

OA pl?
Please think more :-)

OA is different .....Actually I also was puzzled how to reach to that answer....
Sorry Real2008...I missed your 6 in case 1.....

You are indeed correct....can you please tell why are you multiplying the same by 6...why not the solution be as given by "ankitns"...above ?
sorry to poke my nose in..
got the same answer but since u said it was incorrect..i have been thinking it over and over..
the 6 is because you can get 5 heads and 1 tail in 6 ways i.e. 6C5.
man you scared me!!!
The powers of two are bloody impolite!!
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by goelmohit2002 » Thu Aug 06, 2009 10:44 pm
tohellandback wrote: the 6 is because you can get 5 heads and 1 tail in 6 ways i.e. 6C5.
man you scared me!!!
Thanks tohellandback.

But why the same reasoning is not applied in the case of fair coin.....i.e. where the probability is 0.5 ?

i.e. why we do not multiply the same with 6C5 in that ?
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by tohellandback » Thu Aug 06, 2009 10:47 pm
goelmohit2002 wrote:
tohellandback wrote: the 6 is because you can get 5 heads and 1 tail in 6 ways i.e. 6C5.
man you scared me!!!
Thanks tohellandback.

But why the same reasoning is not applied in the case of fair coin.....i.e. where the probability is 0.5 ?

i.e. why we do not multiply the same with 6C5 in that ?
I remember at least one question where this reasoning was applied.
We might help if you can post the question
The powers of two are bloody impolite!!
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by ankitns » Thu Aug 06, 2009 10:53 pm
interesting...so is it the case that we are always supposed to consider the combinations (6C5 in this case) when ever there is an unequal probabilities??

Thanks.
tohellandback wrote:
goelmohit2002 wrote:
goelmohit2002 wrote:
real2008 wrote:case1) five heads and six ways (0.6^5*0.4^1*6)

case2) six heads one way=(0.6^6)

probability= case1+ case2 =0.23328

OA pl?
Please think more :-)

OA is different .....Actually I also was puzzled how to reach to that answer....
Sorry Real2008...I missed your 6 in case 1.....

You are indeed correct....can you please tell why are you multiplying the same by 6...why not the solution be as given by "ankitns"...above ?
sorry to poke my nose in..
got the same answer but since u said it was incorrect..i have been thinking it over and over..
the 6 is because you can get 5 heads and 1 tail in 6 ways i.e. 6C5.
man you scared me!!!
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
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by goelmohit2002 » Thu Aug 06, 2009 10:55 pm
Thanks. Probably I am getting where I am wrong....

The reasoning is applicable to both fair and unfair coin.....

1. HHHHHT - Find how many ways This arrangement is possible....and multiply by the respective head and tail probabiliy....i.e. (6!/(5! *1!)) * (0.6)^5*(0.4)

2. similarly for the case of HHHHHH

3. Finally add 1 and 2.

So I think that the same is applicable for fair coin as well.....

Looks like I was confusing myself with fair and unfair stuff unnecessarily....

Please correct me if I am wrong.....
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by goelmohit2002 » Thu Aug 06, 2009 10:57 pm
ankitns wrote:interesting...so is it the case that we are always supposed to consider the combinations (6C5 in this case) when ever there is an unequal probabilities??

Thanks.
Hi Ankit,

After this discussion, I think we have to consider the cases irrespective of fair/unfair.....

Thanks
Mohit
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by ankitns » Thu Aug 06, 2009 11:11 pm
Yup.
Just tried it with a simple case. 2 tosses on fair coin. What is the probability of atleast 1 head?

Possible options HH, HT, TH, TT

So just by looking at the options above clearly the answer is 3/4.

If we did NOT consider the 2C1 when there is 1 heads we would get 1/4 (for 1 heads) and 1/4 for 2 heads...adding these will give us 1/2..which would be wrong.

But we if did consider the 2C1 when there is 1 head, we would get 1/2 (for 1 heads) and 1/4 (for 2 heads)...adding these will give us 3/4...which is correct.


Thanks.
goelmohit2002 wrote:
ankitns wrote:interesting...so is it the case that we are always supposed to consider the combinations (6C5 in this case) when ever there is an unequal probabilities??

Thanks.
Hi Ankit,

After this discussion, I think we have to consider the cases irrespective of fair/unfair.....

Thanks
Mohit
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
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Re: Head-Tail Question

by adssaini » Fri Aug 07, 2009 5:49 am
ankitns wrote:Case 1 - 5 heads:

(6/10)^5 * (4/10) = (6^5 * 4)/(10^6)


Case 2 - 6 heads = (6^6)/(10^6)

Adding the two cases gives us (6^5)/(10^5)

which can be simplified to (3/5)^5...or (0.6)^5


Whats the OA?

Thanks.

Could you please elaborate how you got (3/5)^5...or (0.6)^5 after adding the two cases?


I added the two of yours and got final answer as .077696 ( .031104 + .046656)

Thanks.
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Re: Head-Tail Question

by ankitns » Fri Aug 07, 2009 7:28 am
Case 1 - 5 heads:

(6/10)^5 * (4/10) = (6^5 * 4)/(10^6)


Case 2 - 6 heads = (6^6)/(10^6)

Adding the two we get

(6^5 * 4)/(10^6) + (6^6)/(10^6)

and then we factor out the common multiples...

= (6^5/10^6) * [4 + 6]

= (6^5/10^6) * [10]

= (6^5/10^5)

= (6/10)^5

= (3/5)^5 or (0.6)^5

Hope this helps.

adssaini wrote:
ankitns wrote:Case 1 - 5 heads:

(6/10)^5 * (4/10) = (6^5 * 4)/(10^6)


Case 2 - 6 heads = (6^6)/(10^6)

Adding the two cases gives us (6^5)/(10^5)

which can be simplified to (3/5)^5...or (0.6)^5


Whats the OA?

Thanks.

Could you please elaborate how you got (3/5)^5...or (0.6)^5 after adding the two cases?


I added the two of yours and got final answer as .077696 ( .031104 + .046656)

Thanks.
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!
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by Ian Stewart » Fri Aug 07, 2009 10:11 am
Going back to the original question, just in case it's unclear why we multiply by 6 when computing the probability of getting 5 Heads and 1 Tails:

The probability of getting, say, the following sequence of Heads and Tails -- HHHHHT -- is equal to (0.6)^5 * (0.4). That is not, however, the only way we might get 5 Heads and 1 Tails; we might have any of the following six sequences:

HHHHHT
HHHHTH
HHHTHH
HHTHHH
HTHHHH
THHHHH

(of course, there are faster ways to count the number of words you can make with 5 H's and 1 T; it's equal to 6C1, for example). The probability of any of the above sequences is equal to (0.6)^5 * (0.4), so adding that number six times, we get the the probability of getting exactly five heads and one tail in some sequence: 6*(0.6)^5 * (0.4).

We need to do the same regardless of whether a coin is fair or unfair.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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