Probability with dice

  • #1
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Homework Statement


If two identical dice are thrown, what is the probability that the total of the numbers is 10 or higher? [Hint: list the combinations that can give a total of 10 or higher.] [2]

Two dice have been thrown, giving a total of at least 10. What is the probability that the throw of a third die will bring the total of the three numbers shown to 15 or higher? [3]

2. Homework Equations ... Actual Answer

Part 1) 1/6
Part 2)
10: 5,6
11: 4,5,6
12: 3,4,5,6

(3/6 x 2/6) + (2/6 x 3/6) + (1/6 x 4/6) = 4/9

The Attempt at a Solution


So I obtained the solution to the first part by drawing a probability space diagram. Giving 1/6

For the second part I thought:
If I get a total of 10 for the first two throws I will need a 5 or 6 to give at least 15
11: 4,5,6
12: 3,4,5,6


P(t=10) = 3/36
P(t=11)= 2/36
P(t=12) = 1/36

2(3/36 x 1/6) + 3(2/36 x 1/6) + 4(1/36 + 1/6) = 2/27

Why am I wrong in assuming that the probability of getting a total of 10 in the first two throughs is the probability represented by the probability space diagram so 3 out of 36. What is wrong with my working?
 

Answers and Replies

  • #2
Orodruin
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You are computing the probability to get at least ten with the first two dice and then a total of at least 15. This is not the same thing as computing the conditional probability to get at least 15 given that the first two dice have a total of at least 10.

Edit: Consider P(A&B) = P(A|B)P(B).
 

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