probability

[anon_econ]

A coin is tossed (m+n) times (m>n). Find the probability of getting at least m consecutive heads.

I have tried this question in several ways but each time I fail to incorporate some cases in my answer. Plz help.

[supriya]

Hi!
at least m consecutive heads => (m,m+1,..... ,m+n)
m heads can be obtained in  n+1 ways
m+1 "      "     "    "          "   n-1+1 "
.                                             .
.                                             .
.                                             .
.
m+n heads can be obtained in n-n+1 ways
summing all these you will get (n+2)(n+1)/2
probability of favorable cases = {[(n+2)(n+1)/2]*(1/2 ^ m+n)}
  Now  you try to workout this one:
If you toss a fair coin m+n times, what is the chance that you get m heads at any time?

[anon_econ]

Hi supriya,
Thanks for ur reply.
But I don't think that answer is complete. For example, if m=6 & n=2, v hv the following favourable cases:
exactly 6 consecutive heads- HHHHHHTT, HHHHHHTH, TTHHHHHH, HTHHHHHH, THHHHHHT
exactly 7 consecutive heads- HHHHHHHT, THHHHHHH
exactly 8 consecutive heads- HHHHHHHH
So, the total number of cases is 8, which is gr8er than (n+1)(n+2)/2.
Ur approach misses out cases like the ones i hv highlighted.

As for the question u hv posted, I think it should be (m+n)Cm*(0.5)^m*(0.5)^n [assuming that u mean exactly m heads]. Is this right?

[duck]

Hi Vasudha.. :)

If n=0, Probability= 1/2^m

If n =1, Probability= 3/2^m+n

If n=2, Probability= 8/2^m+n

If n=3, Probability = 20/2^m+n

and so on...

So, we can write all of the above in the following way>

Probability= n+2/2^m+1

[Amit Goyal]

A coin is tossed (m+n) times (m>n). Find the probability of getting at least m consecutive heads.
This is how you are going to do it. I am going to break the set of favorable events into mutually exclusive events of the type as follows. Figure it out yourself that this takes care of all the possibilities.
Pr(getting at least m consecutive heads) =
Pr(getting consecutive heads in first m tosses) +
Pr(getting tail in the first toss, getting at least m consecutive heads second toss onwards) +
Pr(getting tail in the second toss, getting at least m consecutive heads third toss onwards) +
.
.
.
Pr(getting tail in the nth toss, getting m consecutive heads in last m tosses)
= (0.5)^m + (0.5)^{m+1} +..... + (0.5)^{m+1}  [n-terms of the type (0.5)^{m+1}]
= ((0.5)^m)(1+n(0.5))

[anon_econ]

Thanks a lot sir

[ritu]

sir  i have a doubt in the solution u provided...........
for example....if m=5,n=2.....then in the case THHHHHT.....SHOULDN'T .PR OF 5 HEADS...BE ....(1/2^5)(1/2^2)   and not just (1/2^5)(1/2^1).....means the n cases which have the pr 1/2^m+1.....shouldnt we multiply another 1/2 to it....coz we have to have "m" heads btwn two talis....????

[ritu]

to be more precise what m asking is............in the example @vasudha quoted......if m=6,n=2.....then do we consider" HHHHHHTT "AND "HHHHHHTH "......as two different outcomes or the same?????????????

[duck]

Hi Ritu.. :)

These two are 2 different outcomes.

[Amit Goyal]

If m =  6 and n = 2, then this is what i have done:
Pr(getting at least 6 consecutive heads) =
Pr(getting consecutive heads in first 6 tosses) +
Pr(getting tail in the first toss, getting at least 6 consecutive heads second toss onwards) +
Pr(getting tail in the second toss, getting 6 consecutive heads third toss onwards)
=
Pr({HHHHHHHH, HHHHHHHT, HHHHHHTH, HHHHHHTT}) +
Pr({THHHHHHH, THHHHHHT}) +
Pr({HTHHHHHH, TTHHHHHH})
= (4/(2^8)) + (2/(2^8)) + (2/(2^8))
= (1/(2^6)) + (1/(2^7)) + (1/(2^7))
= ((0.5)^6)(1+(0.5)+(0.5))
= ((0.5)^6)(1+2(0.5))
= (0.5)^5