PRACTICE OF STATS - 1 TERM ACCESS CODE
PRACTICE OF STATS - 1 TERM ACCESS CODE
4th Edition
ISBN: 9781319403348
Author: BALDI
Publisher: Macmillan Higher Education
Question
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Chapter 16, Problem 16.18AT

(a)

To determine

To draw: Tree diagram for the possible outcomes.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information:

    Reaction type to sunNo frecklesFreckles
    I) Always reddens, never tans7973
    II) Always reddens, slight tan581367
    III) Sometimes reddens, always tans1025324
    IV) Never reddens, always tans1022135

Calculation:

Let AR denotes always reddens.

SR denotes sometimes reddens.

NR denotes never reddens.

NT denotes never tans

ST denotes slight tans

AT denotes always tans

F denotes Freckles

NF denotes no freckles

Graph:

Tree diagram

PRACTICE OF STATS - 1 TERM ACCESS CODE, Chapter 16, Problem 16.18AT

(b)

To determine

To compute: The probability of freckles, also, the probability of type 1 reaction to sun exposure.

(b)

Expert Solution
Check Mark

Answer to Problem 16.18AT

  P(freckles(F))=0.2493P(type1reaction)=0.0422

Explanation of Solution

Formula used:

  P(A)=numberoffavourableoutcomestotalnumberofoutcomes

Calculation:

Number of children who had freckles =73+367+324+135

  =899

Total number of children =79+73+581+367+1025+324+1022+135

  =3606

Number of children with type 1 reaction (always reddens, never tans) =79+73

  =152

  P(freckles(F))=8993606=0.2493P(type1reaction)=1523606=0.0422

Thus, probability of a child who had freckles is 0.2493

The probability of type 1 reaction is 0.0422

(c)

To determine

To compute: The probability of type 1 reaction given the child has freckles and vice- verca.

(c)

Expert Solution
Check Mark

Answer to Problem 16.18AT

The probability of type 1 reaction given the child has freckles is 0.0812

The probability of freckles given the child has type 1 reaction =0.4803

Explanation of Solution

Formula used:

Using conditional theorem

  P(A|B)=P(AB)P(B)

Calculation:

Number of children who has type 1 reaction and freckles =73

  P(type1reactionandfreckles)=733606

From the above sub part (b),

  P(type1reaction)=1523606P(freckles)=8993606

   P( type1reaction|freckles )= P( type1reactionfreckles ) P( freckles )

   = 73 3606 899 3606 =0.0812

   P( type1reaction|freckles )= P( type1reactionfreckles ) P( type1reaction )

   = 73 3606 152 3606 =0.4803

Thus, the required probability is 0.0812 and 0.4803

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