An urn contains four balls num

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)