Answer:
29.29
Step-by-step explanation:
The first trial among a distribution of independent trials with a constant success probability follows a geometric probability.
The expected value of a negative binomial is the reciprocal of its success probability p.
E(X)=1/p
At the first draw, we only selected one of the 10 prizes and so 1 draw gives a success
E(X)=1/p
E(X1)=1
After the first prize is drawn, we are interested in the first time (success), that we draw another prize that is different (p=9/10)
E(X2)=1/9÷10
E(X2)= 10/9
After the first prizes have being drawn, we now interested in the first time success that we draw a different prize from the first 2
E(X3)=1/8÷10
E(X3)=10/8
E(X3)=5/4
After the first three prizes have been drawn, we are now interested in the success of a first time draw for the 4th different prize
E(X4)=1/7÷10
E(X4)=10/7
Fifth different prize
E(X5)=1/6÷10
E(X5)=10/6
E(X5)=5/3
Sixth different prize
E(X6)=1/5÷10
E(X6)=10/5
E(X6)=2
Seventh different prize
E(X7)=1/4÷10
E(X7)=10/4
E(X7)=5/2
Eighth different prize
E(X8)=1/3÷10
E(X8)=10/3
Ninth different prize
E(X9)=1/2÷10
E(X9)=10/2
E(X9)=5
After the 9 different prizes have been drawn, we are interested in the first time we will draw the tenth different prize
E(X10)=1/1÷10
E(X10)=10
Add up all corresponding values;
E(X)=E(X1)+E(X2)+E(X3)+E(X4)+E(X5)+E(X6)+E(X7)+E(X8)+E(X9)+E(X10)
E(X)=1+10/9+5/4+10/7+5/3+2+5/2+10/3+5+10
E(X)= 29.29
Note: Those values in E(X), should be written the way it will in standard algebra ie they should be small.
