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

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Manager
Joined: 25 Dec 2011
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GMAT Date: 05-31-2012
A coin is tossed 7 times. Find the probability of getting  [#permalink]

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07 Jan 2012, 13:47
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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

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?

B. 63/128
C. 4/7
D. 61/256

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)
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Posts: 47112

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12 Jan 2012, 05:34
7
6
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))

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:
probability-question-100222.html?hilit=coin%20tossed#p772756
hard-probability-99478.html?hilit=coin%20tossed
some-ps-questions-need-explanation-99282.html?hilit=coin%20tossed
probability-question-gmatprep-85802.html?hilit=coin%20tossed
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Manager
Joined: 25 Dec 2011
Posts: 54
GMAT Date: 05-31-2012

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07 Jan 2012, 13:49
1
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: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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07 Jun 2013, 07:03
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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07 Jun 2013, 08:48
2
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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05 May 2014, 05:47
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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01 May 2015, 02:54
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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01 May 2015, 05:11
2
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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04 Aug 2015, 12:50
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

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?

B. 63/128
C. 4/7
D. 61/256

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.

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

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07 Dec 2016, 20:46
Total outcome = 2^7 = 128

HHHHHHH = 7!/7! = 1
HHHHHHT = 7!/1!6! = 7
HHHHHTT = 7!/2!5! = 21
HHHHTTT = 7!/3!4! = 35

1+7+21+35 = 36
36/128 = 1/2

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

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02 Feb 2017, 04:42
hi bunnel...
As stated in the question" Find the probability of getting more heads than tails in all 7 tosses?"
tails in all the tosses means no heads ie 0 heads and more than that means at least one heads.
so we have to find the the solution for atleast one heads in 7 tosses...
can we restate the question as above?
pls explain

thank u..
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Posts: 47112
Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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02 Feb 2017, 06:12
manojhanagandi wrote:
hi bunnel...
As stated in the question" Find the probability of getting more heads than tails in all 7 tosses?"
tails in all the tosses means no heads ie 0 heads and more than that means at least one heads.
so we have to find the the solution for atleast one heads in 7 tosses...
can we restate the question as above?
pls explain

thank u..

If you read the solutions above you'll see that this is not correct. More heads than tails in all 7 tosses means at least 4 heads (so more than half must be heads):

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

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03 Feb 2017, 04:22
Thank You bunuel...
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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06 Sep 2017, 00:48
Ans is A

HHHHTTT= $$(0.5)^4$$ x $$(0.5)^3$$ {7C4}
similarly for 5 H , 6H and 7 H we will calculate only 7C5 , 7C6 and 7C7 will change other part (0.5)^7 remains same which is multiplied, so add all 4 cases of 4H 5H 6H 7H
$$(0.5)^7$$ x {7C4+7C5+7C6 +7C7}
$$\frac{64}{128}$$= $$\frac{1}{2}$$
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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06 Sep 2017, 03:39
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

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?

B. 63/128
C. 4/7
D. 61/256

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)

Now this is a very simple question and can be solved without doing all the long calculations.
Here important thing to notice is P(H) = P(T) = 1/2 in a single toss.

Now 7 Toss can have following possible ways
0H 7T say x way
1H 6T say y way
2H 5T say z way
3H 4T say w way
--------------------------
4H 3T w way
5H 2T z way
6H 1T x way
7H 0T x way

So 4 out of 8 ways will have heads greater than Tail. Also P(H) = P(T) = 1/2 = i.e. equal in a single toss. So, we don't need to calculate the number of ways for each case.

Require probability = (w+z+x+y)[/(x+y+z+w) +(w+z+x+y)] = 1/2

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

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20 Dec 2017, 09:22
My approach: either Number of heads can be more or less than tails. They can't be equal as no of toss is 7 which is odd.

Hence probability is (no of cases when no of head is more) /((no of cases when no of head is more)+(no of cases when no of head is more))= 1/2

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Bunuel 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

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))

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:
http://gmatclub.com/forum/probability-q ... ed#p772756
http://gmatclub.com/forum/hard-probabil ... n%20tossed
http://gmatclub.com/forum/some-ps-quest ... n%20tossed
http://gmatclub.com/forum/probability-q ... n%20tossed

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A coin is tossed 7 times. Find the probability of getting &nbs [#permalink] 20 Dec 2017, 09:22
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