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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goelmohit2002
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goelmohit2002
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Please think morereal2008 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?
OA is different .....Actually I also was puzzled how to reach to that answer....
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ankitns
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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.
(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!
Attempt 2: Coming soon!
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goelmohit2002
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Sorry Real2008...I missed your 6 in case 1.....goelmohit2002 wrote:Please think morereal2008 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?
OA is different .....Actually I also was puzzled how to reach to that answer....
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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tohellandback
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sorry to poke my nose in..goelmohit2002 wrote:Sorry Real2008...I missed your 6 in case 1.....goelmohit2002 wrote:Please think morereal2008 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?
OA is different .....Actually I also was puzzled how to reach to that answer....
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 ?
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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goelmohit2002
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Thanks tohellandback.tohellandback wrote: the 6 is because you can get 5 heads and 1 tail in 6 ways i.e. 6C5.
man you scared me!!!
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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tohellandback
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I remember at least one question where this reasoning was applied.goelmohit2002 wrote:Thanks tohellandback.tohellandback wrote: the 6 is because you can get 5 heads and 1 tail in 6 ways i.e. 6C5.
man you scared me!!!
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 ?
We might help if you can post the question
The powers of two are bloody impolite!!
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ankitns
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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.
Thanks.
tohellandback wrote:sorry to poke my nose in..goelmohit2002 wrote:Sorry Real2008...I missed your 6 in case 1.....goelmohit2002 wrote:Please think morereal2008 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?
OA is different .....Actually I also was puzzled how to reach to that answer....
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 ?
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!
Attempt 2: Coming soon!
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goelmohit2002
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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.....
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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goelmohit2002
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Hi Ankit,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.
After this discussion, I think we have to consider the cases irrespective of fair/unfair.....
Thanks
Mohit
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ankitns
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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.
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:Hi Ankit,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.
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!
Attempt 2: Coming soon!
Could you please elaborate how you got (3/5)^5...or (0.6)^5 after adding the two cases?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.
I added the two of yours and got final answer as .077696 ( .031104 + .046656)
Thanks.
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ankitns
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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.
(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:Could you please elaborate how you got (3/5)^5...or (0.6)^5 after adding the two cases?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.
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!
Attempt 2: Coming soon!
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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.
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.
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