Question

Felicia has a stack of 10 cards. Each card is labeled with a number in the range 1 through 10 and no two cards have the same number. She picks three cards at random from the stack and adds the numbers on the three cards. What is the expected value of the sum

Answer 1

The expected value of the sum of three numbers chosen out of 10 numbers will be 16.5.

Explanation:

Felicia has a stack of 10 cards with number 1,2,3,4,5,6,7,8,9,10

Selecting 3 cards out of 10

10C3 = 120

Now she picks three cards to look at their numbers and sum it up

The possible three numbers are

(1 2 3), (1 3 2), (2 1 3) ,(2 3 1), (3 1 2), (3 2 1) ,(1 2 4), (1 4 2), (2 1 4), (2 4 1),

(4 1 2)│(4 2 1), (1 2 5), (1 5 2)│(2 1 5), (2 5 1) ,(5 1 2), (5 2 1), (1 2 6), (1 6 2), (2 1 6), (2 6 1), (6 1 2) ,(6 2 1) ,(1 2 7), (1 7 2) ,(2 1 7), (2 7 1), (7 1 2), (7 2 1), (1 2 8), (1 8 2), (2 1 8) ,(2 8 1), (8 1 2) ,(8 2 1), (1 2 9), (1 9 2), (2 1 9), (2 9 1), (9 1 2), (9 2 1),

(1 2... …, (6 10 8), (8 6 10), (8 10 6), (10 6 8), (10 8 6), (6 9 10), (6 10 9), (9 6 10),

(9 10 6), (10 6 9), (10 9 6), (7 8 9), (7 9 8), (8 7 9), (8 9 7), (9 7 8), (9 8 7), (7 8 10),

(7 10 8), (8 7 10), (8 10 7), (10 7 8), (10 8 7), (7 9 10), (7 10 9), (9 7 10), (9 10 7),

(10 7 9), (10 9 7) , (8 9 10), (8 10 9), (9 8 10), (9 10 8), (10 8 9), (10 9 8)

Now we sum it all up so minimum sum can be 1+2+3 = 6 andthe maximum can be 8+9+10 = 27

Let S denote the sum of the three numbers that are chosen and P(S=k) denotes its probability

P(S=6)=1/120

P(S=7)=1/120

P(S=8)=1/60

P(S=9)=1/40

P(S=10)=1/30

P(S=11)=1/24

P(S=12)=7/120

P(S=13)=1/15

P(S=14)=3/40

P(S=15)=1/12

P(S=16)=1/12

P(S=17)=1/12

P(S=18)=1/12

P(S=19)=3/40

P(S=20)=1/15

P(S=21)=7/120

P(S=22)=1/24

P(S=23)=1/30

P(S=24)=1/40

P(S=25)=1/60

P(S=26)=1/120

P(S=27)=1/120

Now adding all the sum in order to get the expected value

∑(K=6 to 27) k*P(S=k) = 33/2 = 16.5

Therefore the expected value of the sum is 16.5.