Question

Appu My Courses - Create Blog Entry pech o s 2. -4.25 POINTS UHSTAT4 8.0.021. MY NOTES A woman decides to have children until she has her first boy or until she has four children, whichever comes first let X = number of chadren she has. For simplicity, assume that the probability of a () The simple events in the simple space are( 0, 000, GGR, GOGG), where we use for boy and G for girl. For instance, one umple event is boy. Find the probability for each of the simple events in the sample space , because the woman quits once she has PGB) POGR) Poda) - PGOGG) (b) Find the probability distribution Munction for X. PEXA)

Answer 1

Solution det & denote the number of children the woman has. Now, P(Boy) = P(Girl) = 0.5 as the two events are independent of each other. a) PCB)= 0.5 PCGB)- P(G).PCB) = 0.5*0.5 = [0:25] PCGGB) = P(G). P(G) • P(8) = 0.5x0.5% 0.5-10.125] P(GGGB) = P(9)•P(G)•P(G)-P(8) = 0.58 0.6X0.50.5-10.0625 P (GGGG ) -P(G) • PCG). PCG). PCG)= 0.5*0.5x0.580570-0625 b) P(x= 1) = P(B) = 0,5 P(x+2) = P(GB) = 0.25 P(x-3) - PC GGB) 20-125 PL X=4) --P@GGB) + P(GGGG) = 0.0625+0.0625 - 0.125 , K 1 2 3 4 P(x x) [65] [ 0- 25] 0-125 | 0 12 S|

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