A basketball player succeeds in making a basket 3 tries out of 4. How many tries arenecessary in order to have probability $\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{99}$of at least one basket?

The probability of successful throw is$\frac{3}{4}$.

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

When events are independent, apply the formula $\mathit{P}\left(AB\right)\mathbf{=}\mathit{P}\left(B\right)\mathbf{·}\mathit{P}\left(B\right)$where andare the events.

The probability of missing the throw is$\frac{1}{4}$. Similarly, the probability of missing n-throws is ${\left(\frac{1}{4}\right)}^n$.

This implies that the probability of at least successful throw is $1-{\left(\frac{1}{4}\right)}^n$.

To have the probability more than$0.99$, $1-{\left(\frac{1}{4}\right)}^n>0.99$.

Solve the obtained inequality to obtain the value of n.

$$\begin{gathered} 1-{\left(\frac{1}{4}\right)}^n>0.99 \\ 0.01>{\left(\frac{1}{4}\right)}^n \\ In\left(0.01\right)>n\ln \left(\frac{1}{4}\right) \\ n<\frac{In\left(0.01\right)}{In\left(0.25\right)} \\ <3.321 \end{gathered}$$

The obtained value is not a whole number; thus we can assume that the required number of throws will be 4.