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According to July \(31, 2013\), posting on cnn.com subsequent to the death of a child who bit into a peanut, a \(2010\) study in the journal Pediatrics found that \(8\% \)of children younger than eighteen in the United States have at least one food allergy. Among those with food allergies, about \(39\% \)had a history of severe reaction.

a. If a child younger than eighteen is randomly selected, what is the probability that he or she has at least one food allergy and a history of severe reaction?

b. It was also reported that \(30\% \) of those with an allergy in fact are allergic to multiple foods. If a child younger than eighteen is randomly selected, what is the probability that he or she is allergic to multiple foods?

Short Answer

Expert verified

a. The probability that he or she has at least one food allergy and a history of severe reaction is \(0.0312\).

b. The probability that he or she is allergic to multiple foods is \(0.024\).

Step by step solution

01

Definition of Probability

Probability is an area of mathematics that deals with numerical descriptions of the likelihood of an event occurring or a proposition being true. An event's probability is a number between \(0{\rm{ }}and{\rm{ }}1\), with zero indicating impossibility and one indicating certainty, broadly speaking.

02

Given Data for part a.

Denote events

\(A = \{ child\;has\;a\;food\;allergy\} ;\)

\(B = \{ child\;has\;a\;history\;of\;severe\;reaction\} \);

We are also given that

\(\begin{array}{l}P(A) &= &0.08\\P(B\mid A) &=& 0.39\end{array}\)

03

Calculation for the determination of probability in part a.

The Multiplication Rule

\(P(A \cap B) = P(A\mid B) \cdot P(B)\)

Using the multiplication rule, we have

\(\begin{array}{c}P(B \cap A) &=& P(B\mid A) \cdot P(A)\\ &=& 0.39 \cdot 0.08\\ &=& 0.0312.\end{array}\)

This is the probability that the child has at least one food allergy and a history of a severe reaction.

04

Given data for part b.

Denote events

\(\begin{array}{l}A = \{ {\rm{ child has a food allergy }}\} ;\\B = \{ {\rm{ child has a history of severe reaction }}\} ;\end{array}\)

We are also given that

\(\begin{array}{l}P(A) &=& 0.08\\P(B\mid A) &=& 0.39\end{array}\)

05

Calculation for the determination for part b.

The Multiplication Rule

\(P(A \cap B) = P(A\mid B) \cdot P(B)\)

Denote event

\(C = \{ a\;child\;younger\;than\;18\;is\;allergic\;to\;multiple\;foods\} \).

We are given that

\(P(C\mid A) = 0.3.\;\;\;(30\% )\)

Again, using the multiplication rule we have

\(P(C)\mathop = \limits^{(1)} P(C \cap A)\mathop = \limits^{(2)} P(A) \cdot P(C\mid A) = 0.08 \cdot 0.3 = 0.024\)

(1): here we use the fact that \(C \subset A\), which indicates that \(C = C \cap A\). This is true because if concept that a person has a food allergy is bigger concept than the person has multiple allergies,

(2): multiplication rule.

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Most popular questions from this chapter

The probability that an individual randomly selected from a particular population has a certain disease is \({\rm{.05}}\). A diagnostic test correctly detects the presence of the disease \({\rm{98\% }}\)of the time and correctly detects the absence of the disease \({\rm{99\% }}\)of the time. If the test is applied twice, the two test results are independent, and both are positive, what is the (posterior) probability that the selected individual has the disease? (Hint: Tree diagram with first-generation branches corresponding to Disease and No Disease, and second- and third-generation branches corresponding to results of the two tests.)

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