Answer:
Step-by-step explanation:
For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6
the probability of a even number is 3/6 = 0.5
Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials
x = number of successful throws
therefore for a Binomial distribution where
P(X =x) = nCx . P^x . (1-P)^ (n-x)
since p = 0.5, and n = 12, the distribution follows
P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)
= 12Cx . 0.5^x . 0.5)^(12- x)
where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)
since we are interested in the probability of the number of times an even number occurs
it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)
For no even number in 12 rolls,
P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244
For one even number in 12 rolls,
P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930
For two even number in 12 rolls,
P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113
For three even number in 12 rolls,
P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711
For four even number in 12 rolls,
P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850
For five even number in 12 rolls,
P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359
For six even number in 12 rolls,
P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586
For seven even number in 12 rolls,
P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359
For eight even number in 12 rolls,
P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850
For nine even number in 12 rolls,
P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711
For ten even number in 12 rolls,
P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113
For eleven even number in 12 rolls,
P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930
For twelve even number in 12 rolls,
P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244
Final test summation[P(X)] = 1
i.e.
P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1
Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,
the probability value stands
