Question

A coin is tossed 6 times . find the probability of getting head in even number (0 is taken as an even

Answer 1

I hope the probability of getting had in even number is 4 ...........

Answer 2

Answer:

The required answer s $\frac{1}{6}$.

Step-by-step explanation:

Given: A coin is tossed $6$ times .

To find: the probability of getting head in even number.

Solution:

Probability of getting head $p=\frac{1}{2}$ and $q=\frac{1}{2}$.

Here $n=6$

P(x=n)=${ }^{\mathrm{n}} \mathrm{Cx} \mathrm{q}^{(\mathrm{n}-\mathrm{x})} \mathbf{p}^{\mathbf{x}}$

Probability of getting $0,2,4,6$ heads respectively are

${ }^{\mathrm{6}} \mathrm{C0} (\frac{1}{2})^{0}(\frac{1}{2})^{6}+{ }^{\mathrm{6}} \mathrm{C2}(\frac{1}{2})^{2}(\frac{1}{2})^{4}+{ }^{\mathrm{6}} \mathrm{C4}(\frac{1}{2})^{4}\((\frac{1}{2})^{2}+{ }^{\mathrm{6}} \mathrm{C6}(\frac{1}{2})^{6}(\frac{1}{2})^{0}$

Substitute the values.

=$1*(\frac{1}{2})^{6}+\frac{6*5}{2}(\frac{1}{2})^{6}+\frac{6*1}{42}(\frac{1}{2})^{6}+1*(\frac{1}{2})^{6}$

Simplify this term.

=$(\frac{1}{2})^{6}(1+15+15+1)$

Solve.

=$\frac{32}{2^{6}}$

=$\frac{1}{6}$

Therefore, the probability of getting head in even number is $\frac{1}{6}$.

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