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A coin is tossed 7 times. Find the probability of getting

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A coin is tossed 7 times. Find the probability of getting [#permalink] New post 07 Jan 2012, 12:47
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A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64

[Reveal] Spoiler:
Hi
I want to understand why combination has been used in the below problem.
I thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.
For example if the question was - probability of getting 1 head then we would have put it as follows :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHHH
TTTTTHH
TTTTTTH

Then why in the below problem - we use combinations and not permutations ?
I am so confused.

A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

A. 1/2 (Correct Answer)
B. 63/128
C. 4/7
D. 61/256
E. 63/64 (Your Answer)

Explanation
ANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)
[Reveal] Spoiler: OA
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Re: probability question [#permalink] New post 07 Jan 2012, 12:49
Hi Sorry ...1 head out of 7 can be arranged in following ways :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHTT
TTTTTHT
TTTTTTH
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Re: probability question [#permalink] New post 12 Jan 2012, 04:34
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A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64

Assuming the coin is fair - P(H)=P(T)=1/2

We can do as proposed by the explanation in your initial post:

Total outcomes: 2^7

Favorable outcomes:
4 heads --> combination of HHHHTTT --> 7!/(4!*3!)=35 (# of permutation of 7 letters out of which 4 H's and 3 T's are identical);
5 heads --> combination of HHHHHTT --> 7!/(5!*2!)=21;
6 heads --> combination of HHHHHHT --> 7!/(6!*1!)=7;
7 heads --> combination of HHHHHHH --> 1;

P(H>T)=Favorable outcomes/Total outcomes=(35+21+7+1)/2^7=1/2.

BUT: there is MUCH simpler and elegant way to solve this question. Since the probability of getting either heads or tails is equal (1/2) and a tie in 7 (odd) tosses is not possible then the probability of getting more heads than tails = to the probability of getting more tails than heads = 1/2. How else? Does the probability favor any of tails or heads? (The distribution of the probabilities is symmetrical: P(H=7)=P(T=7), P(H=5)=P(T=5), ... also P(H>4)=P(T>4))

Answer: A.

If it were: A fair coin is tossed 8 times. Find the probability of getting more heads than tails in all 8 tosses?

Now, almost the same here: as 8 is even then a tie is possible but again as distribution is symmetrical then P(H>T)=\frac{1-P(H=T)}{2}=P(T>H) (so we just subtract the probability of a tie and then divide the given value by 2 as P(H>T)=P(H<T)). As P(H=T)=\frac{8!}{4!*4!}=70 (# of permutation of 8 letters HHHHTTTT, out of which 4 H's and H T's are identical) then P(H>T)=\frac{1-P(H=T)}{2}=\frac{1-\frac{70}{2^8}}{2}=\frac{93}{256}. You can check this in following way: total # of outcomes = 2^8=256, out of which in 70 cases there will be a tie, in 93 cases H>T and also in 93 cases T>H --> 70+93+93=256.

Hope it's clear.

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Re: A coin is tossed 7 times. Find the probability of getting [#permalink] New post 07 Jun 2013, 06:03
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink] New post 07 Jun 2013, 07:48
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Probability of getting Head, P(H) = 1/2

For No. of heads > no. of Tails, there can be 4,5,6 or 7 Heads.

As per GMAT Club math book, probability of occurrence of an event k times in a n time sequence, for independent and mutually exclusive events:

P=C(n,k)*p^k*(1-p)^n^-^k
P(H=4) = C(7,4)*(1/2)^4*(1/2)^3 = C(7,4) * (1/2)^7
P(H=5) = C(7,5) * (1/2)^7
P(H=6) = C(7,6) * (1/2)^7
P(H=7) = C(7,7) * (1/2)^7

Total P(H > 3) = P(H=4) + P(H=5) + P(H=6) + P(H=7)
= [ C(7,4) + C(7,5) + C(7,6) + C(7,7)] * (1/2)^7
= (35 + 21 + 7 + 1) / 128
= 64/128
= 1/2
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink] New post 05 May 2014, 04:47
My approach

7 toss
coin has 2 out comes H or T

now basically if i see this a game where there are 2 teams one selects Heads and other team selects Tails. If in 7 toss whichever comes maximum (heads or Tails) the corresponding team wins..


Clearly the probability is 50% for both the cases max Heads or max Tails............
Re: A coin is tossed 7 times. Find the probability of getting   [#permalink] 05 May 2014, 04:47
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