# An urn contains four balls num

An urn contains four balls numbered 2, 2, 5 and 6. If a personselects a set of two balls at random, what is the expected value ofthe sum of the number of the balls?

Suppose we select two distinct balls

then we have the following possibilities of selecting twoballs

(2,2), (2,2) …………, i.e. 2 ways to get a sum of4……………(2+2 = 4 and 2+2 = 4)

(2,5), (2,5), (5,2), (5,2), i.e. 4 ways to get a sum of7……………(5+2 = 7 and 2+5 = 7)

(2,6), (2,6), (2,6), (2,6), i.e. 4 ways to get a sum of8……………(2+6 = 8 and 6+2 = 8)

(6,5), (5,6) , i.e. 2 ways to get a sum of 11……………(5+6= 11 and 6+5 = 11)

total number of possible outcomes = 2(for a sum of 4) + 4(for asum of 7) + 4(for a sum of 8) + 2 (for a sum of 11)

= 2 + 4 + 4+2

= 12

So, P(getting a sum of 4) = (number of ways to get a sum of4)/total number = 2/12

similarly,

P(getting a sum of 7) = (number of ways to get a sum of 7)/totalnumber = 4/12

P(getting a sum of 8) = (number of ways to get a sum of 8)/totalnumber = 4/12

P(getting a sum of 11) = (number of ways to get a sum of11)/total number = 2/12

We know that expected value E[x] = sum of all individual sumvalues by their respective probability

= 4*(2/12) +7*(4/12) +8*(4/12) +11*(2/12)

= 0.6667 + 2.3333 + 2.6667 + 1.8333

= 7.500 (rounded to 4 decimals)

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