Chapter 10, Problem 10.8AYK

(a)

To determine

To make: a Venn diagram with information and consider options when labeling the diagram.

Answer to Problem 10.8AYK

The Venn diagram was constructed.

Explanation of Solution

Given:

An African study tested 543 children for the sickle-cell anemia trait and for malaria infection. Of those tested, 25% had sickle-cell, 34.6% had malaria and 6.6% had both sickle cell and malaria.

Calculation:

Based on this information a Venn diagram can be constructed. To do this, the probabilities of both malaria and not sickle-cell and sickle-cell and not malaria need to be calculated as follows: $\begin{array}{l}P\left(MandnotS\right)=0.346-0.066\hfill \\ P\left(MandnotS\right)=0.28\hfill \\ P\left(SandnotM\right)=0.25-0.066\hfill \\ P\left(SandnotM\right)=0.184\hfill \end{array}$

To calculate the probability that a child has neither sickle cell nor malaria, first the probability a child has either sickle cell or malaria needs to be determined.

  $\begin{array}{l}P\left(SorM\right)=0.346+0.25-0.066\hfill \\ P\left(SorM\right)=0.53\hfill \end{array}$

From this, the probability that a child has neither sickle cell nor malaria can be calculated by taking 1 minus this above calculated value.

  $\begin{array}{l}P\left(neitherSnorM\right)=1-0.53\hfill \\ P\left(neitherSnorM\right)=0.47\hfill \end{array}$

These values can now be used to complete the Venn diagram.

  

Conclusion:

Therefore, the Venn diagram was constructed.

(b)

To determine

To find: the probability that a given child has either malaria or the sickle-cell trait.

Answer to Problem 10.8AYK

The probability that a child has either malaria or sickle cell is 0.53.

Explanation of Solution

Calculation:

The probability that a given child has neither malaria nor sickle cell can be seen in the Venn diagram. In addition, to calculate the probability that a child has neither sickle cell nor malaria, first the probability a child has either sickle cell or malaria needs to be determined.

  $\begin{array}{l}P\left(SorM\right)=0.346+0.25-0.066\hfill \\ P\left(SorM\right)=0.53\hfill \end{array}$

From this, the probability that a child has neither sickle cell nor malaria can be calculated by taking 1 minus this value.

  $\begin{array}{l}P\left(neitherSnorM\right)=1-0.53\hfill \\ P\left(neitherSnorM\right)=0.47\hfill \end{array}$

The probability that a child has neither sickle cell nor malaria is 0.47.

Also, the probability that a child has either malaria or sickle cell can be determined using the given information. To do this, the probability that a child has sickle cell is added to the probability that a child has malaria. This value is then subtracted from the probability that a child has both.

  $\begin{array}{l}P\left(SorM\right)=0.346+0.25-0.066\hfill \\ P\left(SorM\right)=0.53\hfill \end{array}$

Conclusion:

Therefore,the probability that a child has either malaria or sickle cell is 0.53.

(c)

To determine

To find: the probability that a given child has malaria, given that the child has the sickle-cell trait.

Answer to Problem 10.8AYK

The probability that a child has malaria given they do not have sickle cell is 0.373.

Explanation of Solution

Calculation:

The probability that a child has malaria given that a child has sickle cell is expressed as follows:

  $\begin{array}{l}P\left(M|SC\right)=\frac{P\left(M\text{ and }SC\right)}{P\left(SC\right)}\\ P\left(M|SC\right)=\frac{P\left(0.066\right)}{P\left(0.25\right)}\\ P\left(M|SC\right)=0.264\end{array}$

The probability that a child has malaria given they have sickle cell is 0.264.

The probability that a child has malaria given that they do not have sickle cell is expressed as follows:

  $P\left(M|notSC\right)=\frac{P\left(MandnotSC\right)}{P\left(notSC\right)}$

First, the probability that a child does not have sickle cell needs to be determined by taking 1 minus the probability that a child has sickle cell (0.25).

  $\begin{array}{l}P\left(notSC\right)=1-0.25\\ P\left(notSC\right)=0.75\end{array}$

Next, substitute the values into the formula given above.

  $\begin{array}{l}P\left(M|notSC\right)=\frac{P\left(MandnotSC\right)}{P\left(notSC\right)}\\ P\left(M|notSC\right)=\frac{0.28}{0.75}\\ P\left(M|notSC\right)=0.373\end{array}$

Conclusion:

Therefore,the probability that a child has malaria given they do not have sickle cell is 0.373.

(d)

To determine

To describe: the events sickle-cell trait and malaria independent and might that tell about the relationship between the sickle-cell and malaria.

Answer to Problem 10.8AYK

Therefore, the probability of malaria and sickle cell using the multiplication rule gives a probability of 0.0865.As a result, these two events are not independent.

Explanation of Solution

Calculation:

To verify independence, the multiplication of the probabilities of malaria and sickle cell should be equal to the probability of malaria and sickle cell, which is 0.066.

  $\begin{array}{l}P\left(MandSC\right)=0.066\hfill \\ P\left(MandSC\right)=P\left(M\right)\times P\left(SC\right)\hfill \\ P\left(MandSC\right)=0.346\times 0.25\hfill \\ P\left(MandSC\right)=0.0865\hfill \end{array}$

Calculating the probability of malaria and sickle cell using the multiplication rule gives a probability of 0.0865. This is greater than and does not match the probability of both malaria and sickle cell of 0.066 as given in the example. As a result, these two events are not independent.

Conclusion:

Therefore, the probability of malaria and sickle cell using the multiplication rule gives a probability of 0.0865.As a result, these two events are not independent.