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Chapter 11: Problem 59

Slippery Elum is a baseball pitcher who uses three pitches, \(60 \%\) fastballs, \(25 \%\) curveballs, and the rest spitballs. Slippery is pretty accurate with his fastball (about \(70 \%\) are strikes), less accurate with his curveball (50\% strikes), and very wild with his spitball (only \(30 \%\) strikes). Slippery ends one game with a strike on the last pitch he throws. What is the probability that pitch was a curveball?

Short Answer

Expert verified
The probability that the last pitch was a curveball given that it resulted in a strike is approximately 21.19\%.

Step by step solution

01

Determine the Probability for Each Type of Pitch

We know that Slippery Elum uses three pitches, with the following probabilities:\nFastballs: \(0.60 or 60\% \)\nCurveballs: \(0.25 or 25\% \)\nSpitballs: \(0.15 or 15\% \) - This is because we know that the rest of the pitches are all spitballs. (100% - 60% - 25% = 15%)
02

Determine the Probability of a Strike for Each Type of Pitch

We also know the likelihood of a strike for each type of pitch, which is represented as such:\nFastball strike: \(0.70 or 70\% \)\nCurveball strike: \(0.50 or 50\% \)\nSpitball strike: \(0.30 or 30\% \)
03

Use Bayes' Theorem

Now we can use Bayes鈥� theorem to calculate the exact probability that the last pitch was a curveball given that it was a strike. Bayes' theorem can be used as follows for this problem:\n\(P(Curveball|Strike) = (P(Strike|Curveball) * P(Curveball)) / P(Strike)\)\nThe probability of a strike, P(Strike), can be calculated as follows:\nP(Strike) = P(Strike and Fastball) + P(Strike and Curveball) + P(Strike and Spitball) = (P(Strike|Fastball) * P(Fastball)) + (P(Strike|Curveball) * P(Curveball)) + (P(Strike|Spitball) * P(Spitball))\nUsing the given values, the probabilities become: \nP(Strike) = (0.70 * 0.60) + (0.50 * 0.25) + (0.30 * 0.15) = 0.42 + 0.125 + 0.045 = 0.59. \nNow, we can substitute all the values into Bayes' theorem to find the required probability:\nP(Curveball|Strike) = (0.50 * 0.25) / 0.59 = 0.2119 or 21.19\%.

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Most popular questions from this chapter

To find the proportion of times something occurs, we divide the count (often a binomial random variable) by the number of trials \(n\). Using the formula for the mean and standard deviation of a binomial random variable, derive the mean and standard deviation of a proportion resulting from \(n\) trials and probability of success \(p\).

In a certain board game participants roll a standard six-sided die and need to hit a particular value to get to the finish line exactly. For example, if Carol is three spots from the finish, only a roll of 3 will let her win; anything else and she must wait another turn to roll again. The chance of getting the number she wants on any roll is \(p=1 / 6\) and the rolls are independent of each other. We let a random variable \(X\) count the number of turns until a player gets the number needed to win. The possible values of \(X\) are \(1,2,3, \ldots\) and the probability function for any particular count is given by the formula $$P(X=k)=p(1-p)^{k-1}$$ (a) Find the probability a player finishes on the third turn. (b) Find the probability a player takes more than three turns to finish.

Use the probability function given in the table to calculate: (a) The mean of the random variable (b) The standard deviation of the random variable $$ \begin{array}{lcccc} \hline x & 10 & 12 & 14 & 16 \\ \hline p(x) & 0.25 & 0.25 & 0.25 & 0.25 \\ \hline \end{array} $$

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