Question

calculating probabilities of compound events
a professional pool - player claims that she sinks at least one ball in a pocket on 80% of her opening break shots. in a recent set of 20 games, she sunk at least one ball on her break shot in only 14 of the games. a simulation of 100 trials was conducted to see how unusual these most recent games are.
the professional pool player
based on the dotplot of the simulation results and the pool - players last 20 games, which conclusion can be drawn?
it is most likely that she would sink at least one ball during her break exactly 16 times.
there is about a 27% chance of her sinking at least one ball on her break 14 or fewer times in 20 games. this result is not unusual and there is no convincing evidence that her average is less than 80%.

Explanation:

Step1: Analyze the player's claim

The player claims to sink at least one ball in a pocket on 80% of her opening - break shots. In 20 games, the expected number of games where she sinks at least one ball is \(0.8\times20 = 16\). But just because the expected value is 16 doesn't mean it's most likely to happen exactly 16 times.

Step2: Analyze the simulation results

In the 20 - game sample, she sank at least one ball on her break - shot in 14 games. We need to consider the probability distribution from the simulation. If we assume a binomial - like distribution (since it's a success - failure situation: sink at least one ball or not), we can calculate probabilities. However, from the simulation, we can see that getting 14 or fewer successes out of 20 when the claimed success rate is 80% is not extremely rare.
We calculate the probability of getting 14 or fewer successes in \(n = 20\) trials with \(p=0.8\) using the binomial cumulative distribution function \(P(X\leq14)=\sum_{k = 0}^{14}\binom{20}{k}(0.8)^{k}(0.2)^{20 - k}\approx0.27\) (using a binomial probability calculator or software).

Answer:

There is about a 27% chance of her sinking at least one ball on her break 14 or fewer times in 20 games. This result is not unusual and there is no convincing evidence that her average is less than 80%.