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
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
If its sum of scores is 7 do we have to make possibility space diagram?
I got the solutions of both questions.
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
To calculate the probability that the same score is obtained from both throws, we need to consider the probabilities of obtaining each score on each throw.
We're given that the probabilities of obtaining scores of 2, 3, 4, and 5 are equal. Let's represent this probability as P(X), where X is the score on each throw. Therefore, P(2) = P(3) = P(4) = P(5).
Given: P(6) = 5/9 and P(1) = 1/9.
To find the probability of obtaining the same score on both throws, we need to consider all the possibilities:
1. Both throws result in a score of 6. The probability of this happening is P(6) on the first throw multiplied by P(6) on the second throw, which is (5/9) * (5/9).
2. Both throws result in a score of 1. The probability of this happening is P(1) on the first throw multiplied by P(1) on the second throw, which is (1/9) * (1/9).
3. Both throws result in a score of 2, 3, 4, or 5. The probability of this happening is P(X) on the first throw multiplied by P(X) on the second throw. Since P(2) = P(3) = P(4) = P(5), we can say this probability is P(X) * P(X), which is (P(X))^2.
So, the overall probability of obtaining the same score from both throws is:
(5/9) * (5/9) + (1/9) * (1/9) + (P(X))^2
Since (P(X))^2 represents the probability of obtaining a score of 2, 3, 4, or 5 on both throws, and P(X) is equal for these scores, we can rewrite this as:
(5/9) * (5/9) + (1/9) * (1/9) + 4 * (P(X))^2
Now, we just need to calculate P(X):
P(2) + P(3) + P(4) + P(5) + P(6) + P(1) = 1
Since P(X) is equal for scores of 2, 3, 4, and 5, we can rewrite this as:
4 * P(X) + P(6) + P(1) = 1
Substituting the given values, we have:
4 * P(X) + 5/9 + 1/9 = 1
4 * P(X) + 6/9 = 1
4 * P(X) = 1 - 6/9
4 * P(X) = 3/9
P(X) = (3/9) / 4
P(X) = 1/12
Now we can substitute this value back into the overall probability expression:
(5/9) * (5/9) + (1/9) * (1/9) + 4 * (1/12)^2
Simplifying this expression will give us the final probability of obtaining the same score on both throws.