Statistics WA3: The U.S governors are selected at random without replacement

**The U.S governors are selected at random without replacement**

**Find the probability that the first is a republican and the second a democrat**

Let R represent republican and D represent democratic

P (R1&D1) =P(R1)× (D2/R1)

P (R1) =30/50

P (D2&R1) = 20/49

P (R1 &D1) = P(R1)×(D2&R1)= 30/50×20/49= 0.244

Therefore, the probability that first is republican and the second is a democrat is o.244.

**Find the probability that both are Republican**

P (R1&R2) = P(R1)× (R2/R1)

P (R1) =30/50

P (R2&R1) = 29/49

P (R1 &R2) = P(R1)×(R2&R1)= 30/50×29/49= 0.355

The probability that both are republican is 0.355

**Draw a tree diagram for thus problem similar to the one shown in Fig. 4.25 on page 194**

29/49 R2(R1&R2)= 30/50×29/40= 0.355

R1 20/49 D2 (R1&D2)30/50×20/49= 0.244

30/50

30/49 R2(D1&R2)= 20/50 ×30/49=0.244

20/50

D1 19/49 R2 D1&D2)=20/50×19/49=0.244

**What is the probability that the two governors selected have the same political-party affiliation**

P(R1&R2)+ P(D1&D2)= 0.355+0.155=0.510

**What is the probability that the two governors selected have different political –part affiliation**

P(R1&D2)+ P(D1&D2)= 0.244+0.244=0.490

4.186

a. **Find P(C1)**

P(C1)= 9.3/61.4= 0.151

b. **Find P(C1/S2)**

P(C1/S2)= 13/25.8= 0.050

c. **Are events C1 and S2 independent? Explain your answer**

If P(C1/S2)= P(C1) then events are independent, because 0.050≠0.151 events are dependent

d. **Is the events that an injured person is male independent of the event that an injured person was hurt at home? Explain your answer**.

Male injured person= 35.6

Home injured person= 21.4

P(C2)= 21.4/61.4= 0.348

P(S1)=35.6/61.4=0.579

If P(C2&S2)= P(C2) the events are independent, but because 0.348×0.579≠ 0.348 events are independent.

4.188

a. **find the probability P(G1), P(F1), AND P(G1&F1**).

P(G1)= 0.419/1= 0.419

P(F1)=0.582/1= 0.582

P(G1&F1)=0.300/1=0.300

b. **Use the special multiplication rule to determine whether event G1 and F1 are independent**

If P(G1&F1)=P(G1)×P(F1) when the events are independent, but substituting the values in a. into the equation we see that 0.300±0.419×0.582, so events are depend.

**Reference**

Delucchi, M. (2006). The efficacy of collaborative learning groups in an understanding statistics course. *College teaching, 54*, 244-248.

Muth, J. E. (2006). *Basic Statistics and Pharmaceutical Statistical Application* (Vol. 2). New York: Chapman & Hall. CRC Press.