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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]

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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]

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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]

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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.

Similar questions for practice:
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some-ps-questions-need-explanation-99282.html?hilit=coin%20tossed
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]

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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]

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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]

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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............
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]

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New post 01 May 2015, 01:54
For those who have trouble grasping the concept behind this - check out Khan Academy lessons on probability and combinatorics. I'm not allowed to post urls as a newbie but a simple Google search will throw up the link.

I couldn't make head or tail of these questions before seeing them, tried memorizing the formulae and always messed up. I invested 3 hours in going through those videos and can now solve these questions without knowing any formulae - it's all conceptual. Sal's great with breaking down concepts to simple, relatable stuff.
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A coin is tossed 7 times. Find the probability of getting [#permalink]

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New post 01 May 2015, 04:11
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DipikaP wrote:
For those who have trouble grasping the concept behind this - check out Khan Academy lessons on probability and combinatorics. I'm not allowed to post urls as a newbie but a simple Google search will throw up the link.

I couldn't make head or tail of these questions before seeing them, tried memorizing the formulae and always messed up. I invested 3 hours in going through those videos and can now solve these questions without knowing any formulae - it's all conceptual. Sal's great with breaking down concepts to simple, relatable stuff.


Hi Dipika,

You are totally right when you say "it is all conceptual". Trying to memorise formulae is not the right approach.

For example in this question:
First we should think what are the possible outcomes when we toss a coin: head or tail (2 outcomes)
Now as the coin is fair, the probability that we will get a head or a tail is 1/2

To illustrate, let's take a smaller version of the above question:

What is the probability of getting more heads in 3 tosses?

1st case: We can get 3 heads: HHH
Probability of HHH = 1/2 * 1/2 * 1/2 = 1/8
Probability of getting 3 heads = 1/8

2nd case: We can get two heads: HHT, HTH, THH
Probability of HHT = 1/2 * 1/2 * 1/2 = 1/8
Probability of HTH = 1/2 * 1/2 * 1/2 = 1/8
Probability of THH = 1/2 * 1/2 * 1/2 = 1/8
Probability of getting 2 heads = 3* 1/8

As you can see HHT, HTH and THH are different arrangements of HHT
Probability of getting 2 heads = (No. of arrangements of HHT)* (Probability of getting HHT) = 3!/2! * (1/2 * 1/2 * 1/2) = 3/8

Total probability of getting more heads in 3 tosses = 1/8 + 3/8 = 4/8 = 1/2

Thinking on these lines you can solve the above question easily.

But if you think a little further, we are talking about odd number of tosses (3 and 7). So, either there will be more heads or more tails. It is not possible to get equal number of heads and tails. Hence, in half of the outcomes we will get more heads than tails and in the the other half we will have more tails than heads. Thus, the probability of getting more heads = probability of getting more tails = 1/2.
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]

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New post 04 Aug 2015, 11:50
morya003 wrote:
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)


Since number of times coin has been tossed is 7, either number of heads will be more than tails or vice versa. There is no way number of heads become equal to number of tails. Since head and tails are equally favourable outcome of a coin, possibility of getting more heads = possiblity of getting more tails = 1/2. In other words there are only 2 equally likely events (event 1: heads more than tails, event2: tails more than heads) constituting all the outcomes.

Answer [A]
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Re: A coin is tossed 7 times. Find the probability of getting [#permalink]

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Re: A coin is tossed 7 times. Find the probability of getting   [#permalink] 29 Aug 2016, 22:59
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