A die is thrown 6 times. If 'getting an even number' is a success, find the probability of getting(i) exactly 5 successes(ii) at least 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
(i) exactly 5 successes
(ii) at least 5 successes

A die is thrown 6 times. If ‘getting an even number’ is a success, find the probability of getting
(i) exactly 5 successes
(ii) at least 5 successes

(i) 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 exactly 5 successes will be, either getting 2,4 or 6 i.e., $1 / 6$ probability of each, total, $\frac{3}{6}$ probability, $p=\frac{1}{2}, q=\frac{1}{2}$
The probability of success is $\frac{3}{6}$ and of failure is also $\frac{3}{6}$
Thus, the probability of getting exactly 5 successes will be
$=\frac{\text { The favourable outcomes }}{\text { The total number of outcomes }}$
$\begin{array}{l} \Rightarrow{ }^{6} C_{5}{\frac{3}{6}} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \\ \Rightarrow{ }^{6} C_{5} \frac{1}{64} \\ \Rightarrow \frac{3}{32} \end{array}$
(ii) 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 least 5 successes will be, either getting 2,4 or 6 i.e, $1 / 6$ probability of each, total, $\frac{3}{6}$ probability, $p=\frac{3}{6}, q=\frac{3}{6}$
The probability of success is $\frac{3}{6}$ and of failure is also $\frac{3}{6}$ .
Thus, the probability of getting at least 5 successes will be
$=\frac{\text { The favourable outcomes }}{\text { The total number of outcomes }}$
$\Rightarrow\left({ }^{6} C_{5}+{ }^{6} C_{6}\right) \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6}$
$\begin{array}{l} \Rightarrow\left({ }^{6} \mathrm{C}_{5}+{ }^{6} \mathrm{C}_{6}\right) \frac{1}{64} \\ \Rightarrow \frac{7}{64} \end{array}$