Dice Probability Question?? help (1 Viewer)

[sharoooooo]

Two identical perfect cubes (similar to dice) each having faces numbered 0,1,2,3,4,5 are rolled. A score for the roll is determined as the product of the two numbers on the two upper- most faces.

If the cubes are rolled twice and the scores for each roll are added, what is the probability of a combined score of at least 41?

help me pls
thanks

 

[rand_althor]

The table shows scores for one pair of cubes. The scores which will give a total of at least 41 are (first pair/second pair): 25/25, 20/25 and 25/16.
For 25/25 you need to roll two 5s with both pairs of cubes. The probability of this is (1/6)^2 * (1/6)^2 = 1/1296.
For 20/25 you need to roll a 4 then a 5 with pair 1, and two 5s with pair 2. You can get a 4 and 5 two ways (4 first then 5 or 5 first then 4). This combination (20/25) happen two ways (swap pairs). The probability of this is 2 * 2 * (1/6)(1/6) * (1/6)^2 = 4/1296.
For 25/16, you need to roll two 4s with pair 1, and two 5s with pair 2. This can happen two ways (swap pairs). The probability of this is 2 * (1/6)^2 * (1/6)^2 = 2/1296.

Adding these probabilities: (1+4+2)/1296 = 7/1296.

 

[InteGrand]

Two identical perfect cubes (similar to dice) each having faces numbered 0,1,2,3,4,5 are rolled. A score for the roll is determined as the product of the two numbers on the two upper- most faces.

If the cubes are rolled twice and the scores for each roll are added, what is the probability of a combined score of at least 41?

help me pls
thanks

We just consider the cases that would allow this to happen.

We can't have 25 + 15 (5×5 on one roll and 5×3 on the other), since this is 40 – too small. We can't have anything smaller either. We can have 4×4 + 5×5 though (as this is exactly 16 + 25 = 41).

So the allowable combinations are {((4,4), (5,5)), ((5,5),(5,5)), ((4,5),(5,5))}. (We can observe that anything else, e.g. 4×4 + 5×4 results in a smaller sum than 41).

((4,4), (5,5)) can be obtained from two rolls with a probability of 2×(1/36)×(1/36) (the 2 is there since we could obtain the 4's on the first roll and 5's on the second roll, OR the other way round. The 1/36 is the probability of getting a (4,4) on a given roll.).

((5,5),(5,5)) can be obtained from two rolls with probability 1/(6⁴) (as there is only 1 way to get all 5's, and 6⁴ total possibilities from two rolls).

((4,5),(5,5)) can be obtained from two rolls with probability 4×(1/6)⁴ (since this possibility is equivalent to choosing which die and on which roll will show the 4, with the remaining ones rolling 5; there are 4 ways to choose which die and which roll shows the 4, and for each of these ways, the probability of it occurring is (1/6)⁴).

Hence the required probability P is:

P = 2×(1/36)×(1/36) + 1/(6⁴) + 4×(1/6)⁴

= 2/(6⁴)+ 1/(6⁴) + 4/(6⁴)

= 7/(6⁴)

= 7/1296

≈ 0.00540 (3 sig. fig.).

(Edit: done above already)

 

[sharoooooo]

oh okay
Thanks everyone!!