A die is thrown 6 times. If 'getting an even number' is a success, find the probability of getting at most 5 successes

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A die is thrown 6 times. If ‘getting an even number’ is a success, find the probability of getting at most 5 successes

A die is thrown 6 times. If ‘getting an even number’ is a success, find the probability of getting at most 5 successes

Using Bernoulli’s Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p)$
As the die is thrown 6 times the total number of outcomes will be $6^{6}$ .
And we know that the favourable outcomes of getting at most 5 successes will be, either getting 2,4 or 6 i.e, $1 / 6$ probability of each, total, $\frac{3}{6}$ probability of success.
The probability of success is $\frac{3}{6}$ and of failure is also $\frac{3}{6}$
Thus, the probability of getting at most 5 successes will be
$\begin{array}{l} =\frac{\text { The favourable outcomes }}{\text { The total number of outcomes }} \\ \Rightarrow\left({ }^{6} \mathrm{C}_{0}+{ }^{6} \mathrm{C}_{1}+{ }^{6} \mathrm{C}_{2}+{ }^{6} \mathrm{C}_{3}+{ }^{6} \mathrm{C}_{4}+{ }^{6} \mathrm{C}_{5}\right) \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \\ \Rightarrow\left({ }^{6} \mathrm{C}_{0}+{ }^{6} \mathrm{C}_{1}+{ }^{6} \mathrm{C}_{2}+{ }^{6} \mathrm{C}_{3}+{ }^{6} \mathrm{C}_{4}+{ }^{6} \mathrm{C}_{5}\right) \frac{1}{64} \end{array}$