Asked by Raj
A six side dice is biased so that, when it is thrown, the probability of obtaining a score of 6 is 5/9 and of obtaining scores of 1 is 1/9.The probabilities of obtaining scores of 2,3,4,5 are equal.The dice is thrown twice.Calculate the probability that i.the same score is obtained from both throws
Answers
Answered by
Reiny
to get 6 --- 5/9
to get 1 --- 1/9
to get 2 --- x
....
to get 5 ---- x
4x + 5/9 + 1/9 = 1
solve for x
same score ----> 2 6's, or 2 5's etc
Prob( two 6's ) = (5/9)^2 = 25/81
prob (2 5's) = x^2 <----- you found that above
etc
to get 1 --- 1/9
to get 2 --- x
....
to get 5 ---- x
4x + 5/9 + 1/9 = 1
solve for x
same score ----> 2 6's, or 2 5's etc
Prob( two 6's ) = (5/9)^2 = 25/81
prob (2 5's) = x^2 <----- you found that above
etc
Answered by
Raj
If its sum of scores is 7 do we have to make possibility space diagram?
Answered by
Raj
I got the solutions of both questions.
Answered by
Reiny
For the sum of 7, you could have
1,6 ; 2,5 ; 3,4 ; 4,3 ; 5,2, and 6,1
just calculate the prob of each case, then add them up
e.g. prob (1 and 6) = (1/9)(5/9) = 5/81
prob (3 and 4) = (1/12)(1/12) , you obtained the x from above, hope you had x = 1/12
1,6 ; 2,5 ; 3,4 ; 4,3 ; 5,2, and 6,1
just calculate the prob of each case, then add them up
e.g. prob (1 and 6) = (1/9)(5/9) = 5/81
prob (3 and 4) = (1/12)(1/12) , you obtained the x from above, hope you had x = 1/12
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