Question

2. Using the below table: A A2 0.3 В В 0.4 0.2 0.1 08 a. Compute P(A; or B1). b. Compute P(A) or B2) c. Calculate the marginal probabilities from the following table of joint probabilities. d. Detemine P(A | B1). e. Determine P(A2 B1). f. Did your answers to parts (a) and (b) sum to 1? Is this a coincidence? Explain. g. Calculate P(A; | B2) h. Calculate P(A2| B1). i. Are the events independent? Explain. bivong slde glT8 3. Suppose there are 100 students in your accounting class, 10 of whom are left-handed. Two students are drawn at random. Draw the probability tree and insert the probability for each branch. a. What is the probability of the following events? b. Both are right-handed? Both are left-handed? c. d. One is right-handed and the other is left-handed? At least one is right-handed? е.

Answer 1

PCAJ or B,) P(A1)+P(B,) – P(A&B) =(0.4+0.2)+(0.4+0.3-0.4-0.9| BP(Ajar B3): P(Ar)+P(By) - P(A, 4B2) =(0.4+0.2)+(0.2+0.1)-0.2-10.7 © AT A T A I PIA: 0.4+0.2 10.3+0.1 THAO =0.6 - 0.4 ; , 1В, 0.4+0.3 0.2+ ! PG) -0.7 1 0 .3 @ P(AjIB.) PCA, & B). 0.4 PIB) 0.7 =42=0.57141 © P(A2183) PlA24B.) - 0.3 PCB) - 007 = 3/2 = 0.4286] PA, S B.) +PL Agora 0.9+0.7=1.6 #1 Plasi B), P(Aj & B) P(B) = 0.2%.3= 23=0.6667 P(A2/B.) = P(A&B) = 0.36.7 = 34 =10.4286 OPIA, & B;)= 0.4 PIA,). P(B): 0.6-0.7:0.42 :: PIA,&B,)# P(A) PCB.) Not independent.