A die is thrown 4 times. 'Getting a 1 or a 6 ' is considered a success, Find the probability of getting at most 2 successes

Home » A die is thrown 4 times. ‘Getting a 1 or a 6 ‘ is considered a success, Find the probability of getting at most 2 successes

A die is thrown 4 times. ‘Getting a 1 or a 6 ‘ is considered a success, Find the probability of getting at most 2 successes

A die is thrown 4 times. ‘Getting a 1 or a 6 ‘ is considered a success, Find the probability of getting at most 2 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 $q=(1-p)$
We know that the favourable outcomes of getting at most 2 successes will be, either getting 1 or a 6 i.e, total, $\frac{2}{6}$ probability
The probability of success is $\frac{2}{6}$ and of failure is $\frac{4}{6}$
Thus, the probability of getting at most 2 successes will be
$\begin{array}{l} =\frac{\text { The favourable outcomes }}{\text { The total number of outcomes }} \\ \Rightarrow\left({ }^{4} C_{0}\right) \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{4}{6}+\left({ }^{4} C_{1}\right) \frac{2}{6} \cdot \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{4}{6}+\left({ }^{4} C_{2}\right) \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} \cdot \frac{4}{6} \\ \Rightarrow \frac{72}{81} \\ \Rightarrow \frac{8}{9} \end{array}$