One of ten different prizes was randomly put into each box of a cereal. If a family decided to buy this cereal until it obtained at least one of each of the ten different prizes, what is the expected number of boxes of cereal that must be purchased?

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.