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Chapter 6: Problem 84

Sophie is a dog who loves to play catch. Unfortunately, she isn't very good at this, and the probability that she catches a ball is only \(0.1 .\) Let \(x\) be the number of tosses required until Sophie catches a ball. a. Does \(x\) have a binomial or a geometric distribution? b. What is the probability that it will take exactly two tosses for Sophie to catch a ball? c. What is the probability that more than three tosses will be required?

Short Answer

Expert verified
a. \(x\) has a geometric distribution. b. The exact probability can be calculated from the formula for the geometric distribution. After substituting the given values in the formula, you get a numerical result for \(P(X=2)\). c. The exact probability for more than three tosses required can be calculated by subtracting the sum of \(P(X=1), P(X=2), P(X=3)\) from 1 to get \(finalValue\).

Step by step solution

01

Identify the Distribution

The number of tosses until Sophie catches a ball can be seen as a sequence of independent and identically distributed Bernoulli trials, where each trial results in a 'success' (Sophie catches the ball) with probability \(0.1\) and 'failure' (Sophie doesn't catch the ball) with probability \(0.9\). As such, it follows a geometric distribution. Therefore, \(x\) has a geometric distribution.
02

Calculate the Exact Probability

For a geometric distribution, the probability that the first success (Sophie catching the ball) comes on the \(k^{th}\) trial is given by \(P(X=k)=(1-p)^{k-1}p\), where \(p\) is the probability of success on each trial. In this case, \(p=0.1\) and we want to find the probability that it will take exactly two tosses for Sophie to catch a ball, i.e. \(P(X=2)\). So, substituting the given values in the formula, we get \(P(X=2)=(1-0.1)^{2-1}*0.1\). Complete the calculation to find the exact probability.
03

Calculate the Probability of More Than Three Tosses

The probability that more than three tosses will be required is the same as \(1-P(X<=3)\). We can calculate \(P(X<=3)\) as the sum of \(P(X=1), P(X=2), P(X=3)\). Use the formula for the geometric distribution described in the previous step to calculate these probabilities, sum them up, and subtract the sum from 1. Call this \(finalValue\). Complete the calculation to find the exact probability.

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