A die is weighted so that rolling a 1 is two times as likely as rolling...

A die is weighted so that rolling a 1 is two times as likely as rolling a 2, a 2 is two times likely as rolling a 3, a 3 is two times as likely as rolling a 4, a 4 is two times a likely as rolling a 5, and a 5 is two times as likely as rolling a 6. What is the probability of rolling an even number?

Expert Solution

Lets define,

P(rolling a 6) = k (k is a constant)

P(rolling a 5) = 2*P(rolling a 6) = 2k ( since, rolling a 5 is two times as likely as rolling a 6)

P(rolling a 4) = 2*P(rolling a 5) = 4k ( since, rolling a 4 is two times as likely as rolling a 5)

P(rolling a 3) = 2*P(rolling a 4) = 8k ( since, rolling a 3 is two times as likely as rolling a 4)

P(rolling a 2) = 2*P(rolling a 3) = 16k ( since, rolling a 2 is two times as likely as rolling a 3)

P(rolling a 1) = 2*P(rolling a 2) = 32k ( since, rolling a 1 is two times as likely as rolling a 2)

A die has 6 possible choices 1,2,3,4,5,6

Thus, summing over these six probabilities should lead to 1

P(rolling a 1) + P(rolling a 2) + P(rolling a 3) + P(rolling a 4) + P(rolling a 5) + P(rolling a 6) =1

implies, 32k + 16k + 8k+ 4k + 2k + k = 1

thus,

P(rolling an even number) = P(rolling a 2) +P(rolling a 4) +P(rolling a 6)

milcah answered 1 year ago

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