Probability - 6 coins - occurence of 3 Heads

[gmatrant]

The probability that a certain coin will fall
heads up is 50%. What is the probability that it
falls heads up on three of six tosses?

A) 41/(2^6)
B)35/(2^6)
C)1/(2^6)
D)21/2^5
E)5/(2^4)

How to solve this?

[gmatrant]

please let me know if this is the right answer. I dont have the OA with me.

Total outcomes - 2^6
Favourable outcomes - 3 Heads = 6!/(3!*3!) = 20

Probability = 20/2^6
= 5/(2^4)

[camitava]

Hey gmatrant, dnt mind man but I am not getting why u r calculating like -

Favourable outcomes - 3 Heads = 6!/(3!*3!) = 20

To me, it is coming that the probability will be = 1/2^6. Guys pls focus on this problem!

[gabriel]

5/2^4 for me too ...

.. Gmatrant, I have a request, plz do not post the answers ( or the method) before a few people take a shot at the question .. U can verify your method or answer by checking other poters answer ..

Or, u can use the spoiler , to hide ur post ..

Regards ..

[jangojess]

we need to know the prob of having 3 heads in 6 tosses...
number of cases in which there can be 3 heads in 6 tosses = 6C3 = 20.
prob of having head = prob of having tail = 1/2
prob of having 3 head and 3 tail = (1/2)^6

so prob reqd = 20 * (1/2)^6 = 5/2^4

[gmatrant]

Hey gmatrant, dnt mind man but I am not getting why u r calculating like -

Favourable outcomes - 3 Heads = 6!/(3!*3!) = 20

To me, it is coming that the probability will be = 1/2^6. Guys pls focus on this problem!

Let me explain my approach.. but please correct me if I am wrong..

Each coin can be flipped to Heads or Tails -> so each coin has two possibilities, hence 6 coins have 2*2*2*2*2*2 = 2^6 (Total Outcomes)

Ways of getting 3 Heads, is getting the other 3 Tails.
So I have H,H,H,T,T,T. Now this combination of Head and Tail can be arranged in 6!/(3! *3!) = 20.

So Probability of getting 3 Heads is
Favourable Outcome/ Total Outcome
=> 20/(2^6) = 5/(2^4).

Guys ..lplease et me know if this approach is right.

[samirpandeyit62]

hi gmatrant,
IMO it is good & direct approach, I also solved it like u.

[agni_mba]

there is a straight-forward calc to such problems using binomial distribution:

If p and q are probs. of two outcomes A,B of an event, and this event is repeated n times, then prob. of happening of A (and non-happening of B on such occassions) at exactly 'r' occassions can be represented as:
(nCr)*(p^r)*(q^(n-r))

cheers
nitin