Question

Blood types Each of us has an ABO blood type, which describes whether two characteristics called A and B are present. Every human being has two blood type alleles (gene forms), one inherited from our mother and one from our father. Each of these alleles can be A, B, or O. Which two we inherit determines our blood type. The table shows what our blood type is for each combination of two alleles. We inherit each of a parent’s two alleles with a probability of $0.5$ We inherit independently from our mother and father

(a) Hannah and Jacob both have alleles A and B. Diagram the sample space that shows the alleles that their next child could receive. Then give the possible blood types that this child could have, along with the probability for each blood type.

(b) Jennifer has alleles A and O. Jose has alleles A and B. They have two children. What is the probability that at least one of the two children has blood type B? Show your method.

Short answer

Part (a) The probability is $0.50$

Part (b) The probability is $0.4375$

Step-by-step solution

Part (a) Step 1. Given

The table is:

Part (a) Step 2. Concept

$Probability\left(p\right)=\frac{Numberoffavourable}{Totalnumberofexhaustive}$

Part (a) Step 3. Calculation

The table could be built as follows:

Each blood group's probability can be determined as follows:

$$\begin{gathered} P\left(A\right)=0.25 \\ P\left(B\right)=0.25 \\ P\left(AB\right)=P\left(AB\right)+P\left(AB\right) \\ P\left(AB\right)=0.25+0.25 \\ P\left(AB\right)=0.50 \end{gathered}$$

As a result, a probability of 0.50 is necessary.

Part (b) Step 1. Calculation

B's probability is: $P\left(B\right)=0.25$

Probability does not repeat itself twice.

$$\begin{gathered} P(nottwiceB)=P(B\text{'})P(B\text{'}) \\ =(1-P(B\left)\right)(1-P(B\left)\right) \\ =(1-0.25)(1-0.25) \\ =0.5625 \end{gathered}$$

If Jennifer has alleles A and O and Jose has alleles A and B, the likelihood that at least one of the two offspring has blood type B is computed as follows:

$$\begin{gathered} P(atleastoneB)=1-P(nottwiceB) \\ =1-0.5625 \\ =0.4375 \end{gathered}$$

As a result, $0.4375$ is the required probability.