Question

question provided in picture part a:part b: “assume the selections are made without replacement what is the probability that the 2 selected subjects are both group B and type Rh+?” (round to four decimal places as needed)

Answer 1

The probability of picking a sample that is from group B, and type Rh+ is

$\begin{gathered} P(B\text{ and }Rh^+)=\frac{\text{B and Rh}^+}{\text{Total number of samples}} \\ P(B\text{ and }Rh^+)=\frac{23}{60+57+23+23+12+12+6+3} \\ P(B\text{ and }Rh^+)=\frac{23}{196} \end{gathered}$

Part A:

Given that the selection are made with replacement, the probability that the 2 selected subjects are both group B and type Rh+ is

$\begin{gathered} P(B\text{ and }Rh^+\text{ two times with replacement})=P(B\text{ and }Rh^+)\cdot P(B\text{ and }Rh^+) \\ P(B\text{ and }Rh^+\text{ two times with replacement})=\frac{23}{196}\cdot\frac{23}{196} \\ P(B\text{ and }Rh^+\text{ two times with replacement})=0.01377030404 \end{gathered}$

Rounding our answer to four decimal place, the probability is 0.0138.

Part B:

If without replacement is made within the selection, the probability is changed so that

$\begin{gathered} P(B\text{ and }Rh^+\text{ two times})=\frac{\text{B and Rh}^+}{\text{Total number of samples}}\cdot\frac{\text{B and Rh}^{+\text{ }}\text{left}}{\text{number of samples left}} \\ P(B\text{ and }Rh^+\text{ two times})=\frac{23}{196}\cdot\frac{22}{195} \\ P(B\text{ and }Rh^+\text{ two times})=\frac{506}{38220} \\ P(B\text{ and }Rh^+\text{ two times})=0.01323914181 \end{gathered}$

Rounding to answer to four decimal place, the probability is 0.0132