# Probability and Sample Space One particular high school encourages student to donate blood. The high school gym is set up for this purpose. The distribution of blood types in North...

** Probability and Sample Space**

**One particular high school encourages student to donate blood. The high school gym is set up for this purpose. The distribution of blood types in North America is as follows:**

**Type O: 44%**

**Type A: 42%**

**Type B: 10%**

**Type AB: 4%**

**a) What is the probability that the first two people in the line up have the same type of blood? Express your answer to the nearest hundredth of one percent. **

**b) What is the probability that none of the first five people in the line up have type AB blood? Express your answer to the nearest hundredth of one percent. **

### 1 Answer | Add Yours

**a)** ` `

Let AA = first two people have blood type A, BB = first two people have blood type B, etc. You are looking for probability

`P(X in {\A\A cup BB cup OO cup ABAB})=`

Since those are mutually exclusive events (e.g. if first two persons have blood type A, then they cannot have blood type B) probability of their union is equal to sum of their probabilities.

`=P(\A\A)+P(BB)+P(OO)+P(ABAB)`

Since probability of ones persons blood type does not depend upon other persons blood type we have probability of independent events, hence

`P(\A\A)=`probability that first person has blood type A times probability that second person has blood type A`=0.42cdot0.42`. Hence

`P(\A\A)+P(BB)+P(OO)+P(ABAB)=`

`0.42^2+0.1^2+0.44^2+0.04^2=0.3816=38.16%`

**b)**

Probability of not having blood type B is probability of opposite event thus probability that a person doesn't have blood type B is

`P(X ne B)=1-P(B)=1-0.04=0.96=96%`

Since blood types of first five people are mutually independent we have

`P(BBBBB)=0.96 cdot 0.96 cdot 0.96 cdot 0.96 cdot 0.96=0.96^5`

`P(BBBBB)approx0.8154=81.54%`