A basketball player succeeds in making a basket 3 tries out of 4 . How many tries are necessary in order to have probability \(>0.99\) of at least one basket?

Probability theory is the branch of mathematics that deals with the analysis of random events. The core idea is to quantify the likelihood of different outcomes.
When dealing with probability, we often talk about the chance of an event occurring. This can be expressed as a number between 0 and 1, where 0 means the event never happens, and 1 means it always happens. For example, a probability of 0.75 means there's a 75% chance of the event happening.
In the given exercise, we have a basketball player who makes a basket with a probability of 0.75. Here, 0.75 is the probability of success. Conversely, the probability of not making a basket, or failure, is 1-0.75 = 0.25, or 25%.

The complementary probability is the probability that the complementary event will occur. If we know the probability of an event happening, we can find the probability of it not happening by subtracting the probability of the event from 1.
In our example, the probability of the player making at least one basket is the complementary probability of not making any baskets. If the probability of making zero baskets in n tries is \(0.25^n\), then the probability of making at least one basket is the complement: \(1 - 0.25^n\).
Using complementary probability allows us to turn complex problems into simpler ones by focusing on the 'non-event.'

Inequalities are mathematical expressions involving the symbols \(<, \, >, \, \, \leq\, or \geq\) which describe the relative size of two values. In probability, inequalities help us set up conditions to solve problems.
In our exercise, we need the probability of making at least one basket to be more than 0.99. This leads us to the inequality:
\[1 - 0.25^n > 0.99\].
By rearranging this, we get:
\[0.25^n < 0.01\].
This inequality tells us how many tries are needed to achieve a high probability of success.

Natural logarithms (ln) are the logarithms to the base e, where e is approximately equal to 2.71828. Logarithms help us solve equations where the variable is an exponent.
In our problem, to solve the inequality \[0.25^n < 0.01\], we take the natural logarithm of both sides:
\[\text{ln}(0.25^n) < \text{ln}(0.01)\].
Using the property of logarithms \((\text{ln}(a^b) = b \times \text{ln}(a))\), this simplifies to:
\[n \times \text{ln}(0.25) < \text{ln}(0.01)\].
Since \text{ln}(0.25)\ is negative, dividing both sides by it flips the inequality:
\[n > \frac{\text{ln}(0.01)}{\text{ln}(0.25)}\].
Finally, calculating this gives us the number of tries needed to surpass the 0.99 probability threshold.