A fair cubical die is thrown twice and their scores summed up. If the sum of the scores of upper side faces by throwing two times a die is an event. Find the Expected Value of that event.(a) 48(b) 76(c) 7(d) 132I have been asked this question during an internship interview.My question is taken from Discrete Probability topic in division Discrete Probability of Discrete Mathematics

A fair cubical die is thrown twice and their scores summed up. If the

1.	A fair cubical die is thrown twice and their scores summed up. If the sum of the scores of upper side faces by throwing two times a die is an event. Find the Expected Value of that event.(a) 48(b) 76(c) 7(d) 132I have been asked this question during an internship interview.My question is taken from Discrete Probability topic in division Discrete Probability of Discrete Mathematics
Answer» Correct choice is (c) 7 The best I can explain: SAMPLE space = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.Suppose: P(2) = \(\frac{1}{36}\), P(3) = \(\frac{2}{36}\), P(4) = \(\frac{3}{36}\), P(5) = \(\frac{4}{36}\), P(6) = \(\frac{5}{36}\), P(7) = \(\frac{6}{36}\), P(8) = \(\frac{5}{36}\), P(9) = \(\frac{4}{36}\), P(10) = \(\frac{3}{36}\), P(11) = \(\frac{2}{36}\) and P(12) = \(\frac{1}{36}\). Now, Expected VALUE: μ = E(A) = ∑x * P(x) = \(2 * \frac{1}{36} + 3 * \frac{2}{36} + 4 * \frac{3}{36} + 5 * \frac{4}{36} + 6 * \frac{5}{36} \) \(+ 7 * \frac{6}{36} + 8 * \frac{5}{36} + 9 * \frac{4}{36} + 10 * \frac{3}{36} + 11 * \frac{2}{36} + 12 * \frac{1}{36} = \frac{252}{36}\) = 7.

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