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Chapter 7: Problem 38

Based on data obtained from the National Institute of Dental Research, it has been determined that \(42 \%\) of 12 -yr-olds have never had a cavity, \(34 \%\) of 13 -yr-olds have never had a cavity, and \(28 \%\) of 14 -yrolds have never had a cavity. Suppose a child is selected at random from a group of 24 junior high school students that includes six 12 -yr-olds, eight 13 -yr-olds, and ten 14 -yrolds. If this child does not have a cavity, what is the probability that this child is 14 yrs old?

Short Answer

Expert verified
The probability that the child is 14 years old, given that they do not have a cavity, is approximately \(0.306\).

Step by step solution

01

Calculate the probabilities of selecting each age group

First, we calculate the probabilities of selecting a 12-yr-old, 13-yr-old, and 14-yr-old child. There are 24 children in total, with six 12-yr-olds, eight 13-yr-olds, and ten 14-yr-olds. Using these values, Probability of selecting a 12-yr-old: \(P(A_1) = \frac{6}{24}\) Probability of selecting a 13-yr-old: \(P(A_2) = \frac{8}{24}\) Probability of selecting a 14-yr-old: \(P(A_3) = \frac{10}{24}\)
02

Calculate the probabilities of not having a cavity for each age group

We are given the probabilities of not having a cavity for each age group: For 12-yr-olds: \(P(B|A_1) = 0.42\) For 13-yr-olds: \(P(B|A_2) = 0.34\) For 14-yr-olds: \(P(B|A_3) = 0.28\)
03

Calculate the probability of selecting a child without a cavity

For this, we can use the total probability formula, which states that \(P(B)= P(A_1)P(B|A_1) + P(A_2)P(B|A_2) + P(A_3)P(B|A_3)\) So, \(P(B)= \frac{6}{24}(0.42) + \frac{8}{24}(0.34) + \frac{10}{24}(0.28)\)
04

Apply Bayes' theorem to calculate the probability of a 14-year-old child given they don't have a cavity

Using Bayes' theorem, the probability of selecting a 14-year-old given that the child doesn't have a cavity is: \(P(A_3|B) = \frac{P(A_3) P(B|A_3)}{P(B)}\) Plug in the values that we obtained in steps 1, 2, and 3: \(P(A_3|B) = \frac{\frac{10}{24}(0.28)}{\frac{6}{24}(0.42) + \frac{8}{24}(0.34) + \frac{10}{24}(0.28)}\) Finally, after computing the above expression, you'll get the probability that the child is 14 years old: \(P(A_3|B) \approx 0.306\)

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Most popular questions from this chapter

Refer to the following experiment: Urn A contains four white and six black balls. Urn B contains three white and five black balls. A ball is drawn from urn A and then transferred to urn B. A ball is then drawn from urn B. Represent the probabilities associated with this two-stage experiment in the form of a tree diagram.

A coin is tossed three times. What is the probability that the coin will land heads a. At least twice? b. On the second toss, given that heads were thrown on the first toss? C. On the third toss, given that tails were thrown on the first toss?

In a survey of 106 senior information technology and data security professionals at major U.S. companies regarding their confidence that they had detected all significant security breaches in the past year, the following responses were obtained. $$\begin{array}{lcccc} \hline & \begin{array}{c} \text { Very } \\ \text { confident } \end{array} & \begin{array}{c} \text { Moderately } \\ \text { confident } \end{array} & \begin{array}{l} \text { Not very } \\ \text { confident } \end{array} & \begin{array}{l} \text { Not at all } \\ \text { confident } \end{array} \\ \hline \text { Respondents } & 21 & 56 & 22 & 7 \\ \hline \end{array}$$ What is the probability that a respondent in the survey selected at random a. Had little or no confidence that he or she had detected all significant security breaches in the past year? b. Was very confident that he or she had detected all significant security breaches in the past year?

Refer to the following experiment: Urn A contains four white and six black balls. Urn B contains three white and five black balls. A ball is drawn from urn A and then transferred to urn B. A ball is then drawn from urn B. What is the probability that the transferred ball was black given that the second ball drawn was white?

A study conducted by the Metro Housing Agency in a midwestern city revealed the following information concerning the age distribution of renters within the city. $$\begin{array}{lcc} \hline & & \text { Group } \\ \text { Age } & \text { Adult Population, \% } & \text { Who Are Renters, \% } \\\ \hline 21-44 & 51 & 58 \\ \hline 45-64 & 31 & 45 \\ \hline 65 \text { and over } & 18 & 60 \\ \hline \end{array}$$ a. What is the probability that an adult selected at random from this population is a renter? b. If a renter is selected at random, what is the probability that he or she is in the \(21-44\) age bracket? c. If a renter is selected at random, what is the probability that he or she is 45 yr of age or older?
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