A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has

Question

Mathematics

Ayaan Vazquez

2 April, 01:36

A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has

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Answers (1)

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Malia Morrison · Answer 1

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