17. A pair of dice is thrown together, find the expected value.
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Answer:
When two fair die are rolled, the number of possible outcomes = 6 × 6 = 36.
Let X be the sum of the numbers, this can range from 2 to 12.
P (X = 2) = P {(1,1)} = 136
P (X = 3) = P {(1,2), (2,1)} = 236
P (X = 4) = P ({(1,3), (3,1), (2,2)} = 336
P (X = 5) = P ({(1,4), (4,1), (2,3), (3,2} = 436
P (X = 6) = P ({(1,5), (5,1), (4,2), (2,4), (3,3)} = 536
P (X = 7) = P {(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)} = 636
P (X = 8) = P {(2,6), (6,2), (3,5), (5,3), (4,4)} = 536
P (X = 9) = P {(3,6), (6,3), (4,5), (5,4)} = 436
P (X = 10) = P {(4,6), (6,4), (5,5)} = 336
P (X = 11) = P {(5,6), (6,5)} = 236
P (X = 12) = P {(6,6)} = 136
Mean of the probability distribution = ∑(Xi×P(Xi))
Mean or E(X)=∑(Xi×P(Xi)) = 2×136+3×236+4×336+5×436+6×536+7×636+8×536+9×436+10×336+11×236+12×136
Mean = 4+6+12+20+30+42+40+36+30+22+1236=25236=7
Standard Deviation = Variance−−−−−−−√, where Variance =E(X2)−E(X)2
E(X2)=∑((Xi)2×P(Xi)) = 22×136+3×236+42×336+52×436+62×536+72×636+82×536+92×436+102×336+112×236+122×136
E(X2)=4+18+48+100+180+294+320+324+300+242+14436=197436=54.83
Therefore Variance = 54.83−72=54.83−49=5.83
Therefore Standard Deviation = 5.83−−−−√=2.41
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