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A coin is tossed 4 times. What is the probability that the

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A coin is tossed 4 times. What is the probability that the [#permalink]

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A coin is tossed 4 times. What is the probability that the number of Heads is equal to the number of Tails?

A. 1/8
B. 1/4
C. 3/8
D. 1/2
E. 9/16

help!
[Reveal] Spoiler: OA
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Re: A coin is tossed 4 times [#permalink]

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New post 18 Jan 2012, 02:39
We have to find the probability that we get 2 heads and 2 tails in 4 tosses of the coin.

Using the binary formula and defining a head as a success,
P(2 heads in 4 tosses) = 4C2* (1/2)^2 * (1/2)^2
= 6/16
= 3/8

The answer is therefore (C)
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Re: A coin is tossed 4 times [#permalink]

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New post 18 Jan 2012, 02:59
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Splendidgirl666 wrote:
A coin is tossed 4 times. What is the probability that the number of Heads is equal to the number of Tails?

1. 1/8
2. 1/4
3. 3/8
4. 1/2
5. 9/16

help!


Solution:
We need the probability of: HHTT.

\(P(HHTT)=\frac{4!}{2!*2!}*(\frac{1}{2})^2*(\frac{1}{2})^2=\frac{3}{8}\).

Explanation:
\(\frac{4!}{2!*2!}=6\) is number of all possible cases for our wining scenarios HHTT to appear (which is number of permutation of 4 letters HHTT where 2 H's and 2T's are identical: HHTT, HTHT, HTTH, TTHH, THTH, THHT), \((\frac{1}{2})^2\) is the probability of HH and \((\frac{1}{2})^2\) is the probability of TT.

Theory behind the concept above:
Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

General theory on this topic:
Probability: math-probability-87244.html
Combinatorics: math-combinatorics-87345.html

Hope it helps.
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Re: A coin is tossed 4 times [#permalink]

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New post 28 Feb 2012, 10:41
+1 C

First, we have to calculate the total ways of organizing HHTT. (H: Head; T: Tail):
\(\frac{4!}{2!2!} = 6\)

In any of these ways, the probability is:

\(\frac{1}{2} * \frac{1}{2} * \frac{1}{2} * \frac{1}{2} = \frac{1}{16}\)

So, we multiply that probability by the number of ways (6):

\(\frac{1}{16} * 6 = \frac{6}{16} = \frac{3}{8}\)

Answer: C
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Re: A coin is tossed 4 times. What is the probability that the [#permalink]

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New post 10 Jul 2016, 04:52
Heads equal to tails would be : 2 heads and 2 tails.
i.e. HHTT
4!/2! 2! = 6
6/ 2^4 = 3/8
Ans. C
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Re: A coin is tossed 4 times. What is the probability that the [#permalink]

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Re: A coin is tossed 4 times. What is the probability that the   [#permalink] 07 Jan 2018, 09:37
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