Question

At the carnival, we found a bean bag toss booth. There's a 0.66 probability of throwing the bean bag into the outer ring, a 0.17 probability of throwing it into the middle ring, and the remaining probability throwing it into the center of the target. Tickets are awarded depending on which ring the bean bag lands in. We get 1 ticket for the outer ring, 2 tickets for the middle ring, and 3 tickets for the center. What is the expected value (number of tickets) we will get if we play the game 10 times?

	Outer Ring	Middle Ring	Center Ring
Probability	0.66	0.17
Tickets	1	2	3

Answer 1

let X be number of tickets from 1 game

P(X=1) =0.66

P(X=2) =0.17

P(X=3) =1-0.66-0.17 =0.17

therefore expected tickets from 1 game =E(x)=xP(x )=0.66*1+2*0.17+3*0.17

=1.51

hence expected tickets from 10 game =1.51*10 =15.1