Chapter 16, Problem 16.18AT

(a)

To determine

To draw: Tree diagram for the possible outcomes.

Explanation of Solution

Given information:

Reaction type to sun	No freckles	Freckles
I) Always reddens, never tans	79	73
II) Always reddens, slight tan	581	367
III) Sometimes reddens, always tans	1025	324
IV) Never reddens, always tans	1022	135

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

(b)

To determine

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

Answer to Problem 16.18AT

  $\begin{array}{l}P\left(freckles\left(F\right)\right)=0.2493\\ P\left(type1reaction\right)=0.0422\end{array}$

Explanation of Solution

Formula used:

  $P\left(A\right)=\frac{numberoffavourableoutcomes}{totalnumberofoutcomes}$

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) $=79+73$

  $=152$

  $\begin{array}{l}P\left(freckles\left(F\right)\right)=\frac{899}{3606}\\ =0.2493\\ P\left(type1reaction\right)=\frac{152}{3606}\\ =0.0422\end{array}$

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.

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

  $P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$

Calculation:

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

  $P\left(type1reactionandfreckles\right)=\frac{73}{3606}$

From the above sub part (b),

  $\begin{array}{l}P\left(type1reaction\right)=\frac{152}{3606}\\ P\left(freckles\right)=\frac{899}{3606}\end{array}$

  $P\left(type1reaction|freckles\right)=\frac{P\left(type1reaction\cap freckles\right)}{P\left(freckles\right)}$

  $\begin{array}{l}=\frac{\frac{73}{3606}}{\frac{899}{3606}}\\ =0.0812\end{array}$

  $P\left(type1reaction|freckles\right)=\frac{P\left(type1reaction\cap freckles\right)}{P\left(type1reaction\right)}$

  $\begin{array}{l}=\frac{\frac{73}{3606}}{\frac{152}{3606}}\\ =0.4803\end{array}$

Thus, the required probability is $0.0812$ and $0.4803$