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igor_vitrenko [27]
3 years ago
15

Four (4) fair dice are rolled. what is the probability that the total is 21

Mathematics
1 answer:
liberstina [14]3 years ago
4 0
Short Answer 5/216
Comment
The question really is, how many different types of throws with 4 dice will give 21? You can count them

Pattern 1
6663 

Pattern 2
5556

Pattern 3 [ The hard one]
6654

Patterns 1 and 2 give 4 each.
6 6 6 3
6 6 3 6
6 3 6 6
3 6 6 6

5553 will do the same thing

Both can be found by (4/1) as a combination. 

Now for 6654 That gives 12

6 6 5 4
6 6 4 5
6 5 4 6
6 5 6 4
6 4 5 6
6 4 6 5
5 6 4 6
5 6 6 4
5 4 6 6
4 5 6 6
4 6 5 6
4 6 6 5

That should total 12 
The total number of ways of getting 21 with this pattern is 12

Total successes
12 + 4 + 4 = 30

What is the total number of ways you can throw 4 dice?
Total = 6 * 6 * 6 * 6 = 1296

What is the probability of success?
30 / 1296 = 5 / 216
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A pair of standard dice is rolled find the probability that the sum of the two dice is greater than 12
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Probability that the sum of the two dice is greater than 12 is ZERO.

Step-by-step explanation:

Here, when two dices are rolled together, the sample space is given as:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,6)

(6, 1), (6, 2), (6, 3), (6, 4),(6,5), (6,6)  = Total 36 outcomes

Now, E :  Event of getting a  sum greater than 12.

So here, number of possible outcomes  = 0

So, the probability that the sum of the two dice is greater than 12  = 0/36  = 0

So, it is an IMPOSSIBLE EVENT.

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