# A coin is flipped 12 times. What is the probability of getting exactly 8 heads?

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6
Nov 3, 2016

$\frac{495}{4096}$

#### Explanation:

The number of possible sequences of heads and tails in $12$ coin tosses is:

${2}^{12} = 4096$

The number of ways that such a sequence could contain exactly $8$ heads is the number of ways of choosing $8$ out of $12$...

((12),(8)) = (12!)/((8!)(4!)) = (12xx11xx10xx9)/(4xx3xx2xx1) = 495

So the probability of exactly $8$ heads in $12$ coin tosses is:

$\frac{495}{4096}$

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1
Nov 4, 2016

Probability of getting exactly 8 heads in tossing a coin 12 times is $\frac{495}{4096.}$

#### Explanation:

If a coin is tossed 12 times, the maximum probability of getting heads is 12.
But, 12 coin tosses leads to ${2}^{12}$, i.e. 4096 number of possible sequences of heads & tails.

Let E be an event of getting heads in tossing the coin and S be the sample space of maximum possibilities of getting heads. Then probability of the event E can be defined as,

$P \left(E\right) = \frac{n \left(E\right)}{n \left(S\right)} .$ [Where, n represents number accordingly].

$\therefore P \left(E\right) = \frac{495}{4096.}$ (answer).

http://www.careerbless.com/aptitude/qa/probability_imp.php.

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