Question

Imagine taking a lottery ticket. You randomly draw a ticket from a bucket of ticket where at the time you attend there are only 30 tickets left. Of these 30 tickets, 3 are winning tickets. You withdraw 5 lots (without putting them back in the bucket). Let X be the number of winning lots among the 5 drawers you draw.
a) What distribution does X have? Explain.
Calculate P (X = 1), P (X ≥ 1) and E (X).
Now imagine that you are offered to draw the 5 tickets with the cover. That is, after you pull out one ticket and check if it is a winning ticket or not, put it back in the bucket before doing the next draw. Suppose that the bucket is shaken between each draw and that you should not look down into the bucket when doing the draws so that the draws become independent.
b) What distribution does X now have? Explain.
Calculate what P (X = 1), P (X ≥ 1) and E (X) are now.

Answer 1

8.30 tickets: 3 winning 27 non winning @ P(x-1)= P(choosing 1 from 3 and 4 from 27) (9 - 0.3695 "(*1): 1-PCX=0)='s-P(choosing cell 5 from 27) non winning tickets) - 0.4335 ***P(x=x)= () ET) ; =0,1,2,3 - Ex)= x2(2 ) } <)Em) 71253 142506 (0+52 650 + 17550 +1053) - 72506 - 10.5 ina & have hyper geometric distribution. Alternate ways of Ecx) = 5x Zo = 0.5 6 x follows binomial dist where n=5, p. 320.00 P(x=x) = (2) P² (1-pin- x = (5) 60.1 (0.gs.x P(x-1)=(5)10.1 (0.99= 0.32805 P(x1) = 1-PCX=0)1-(5)(0.1°(0.9)% = 0.40951 5(x) = m = 5* 0.1 = 0.51