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anonymous
 one year ago
A red 20sided die, with faces marked from 1 to 20 , is rolled. The result is denoted by D1 . Then a green 20sided die, similarly marked, is rolled. The result here is denoted by D2 .
1) What is the probability that the value 6 shows up on at least on of the rolls? P(D1=6∪D2=6)
2) What is the probability that the value 6 show up on exactly one of the rolls? P((D1=6∪D2=6)∖(D1=6∩D2=6))
3) What is the probability (as a rational number is a/b format) that the value sum D1+D2 is 30 ? 10 ? 23 ?
P(D1+D2=30)
P(D1+D2=10)
P(D1+D2=23)
anonymous
 one year ago
A red 20sided die, with faces marked from 1 to 20 , is rolled. The result is denoted by D1 . Then a green 20sided die, similarly marked, is rolled. The result here is denoted by D2 . 1) What is the probability that the value 6 shows up on at least on of the rolls? P(D1=6∪D2=6) 2) What is the probability that the value 6 show up on exactly one of the rolls? P((D1=6∪D2=6)∖(D1=6∩D2=6)) 3) What is the probability (as a rational number is a/b format) that the value sum D1+D2 is 30 ? 10 ? 23 ? P(D1+D2=30) P(D1+D2=10) P(D1+D2=23)

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Jadedry
 one year ago
Best ResponseYou've already chosen the best response.21) If the value 6 appears on at least one of the dice, then it would be easier to calculate the probability of 6 not appearing at all. 19/20 * 19/20 = 361/400 1 361/400 = 39/400 2) If 6 only shows up on one of the dice, this leaves two possibilities. D1 = 6 and D2 = other, or D1 = other and D2 = 6 (19/20 * 1/20) + (1/20 * 19/20) = 38/400 3) The numbers that equal 30 together = 20 , 10 19 , 11 18 , 12 17, 13 16,14 15,15 14,16 13,17 12,18 11,19 10, 20 = 11 possibilities in each case the probability of getting those 2 numbers = 1/20 * 1/20 = 1/400 = 1/400 * 11 = 11/400  For number 10 we have: 1, 9 2,8 3,7 4,6 5,5 4,6 3,7 2,8 1,9 = 9 possibilities = (1/20 * 1/20) * 9 = 9 /400  For 23 = 20 , 3 19, 4 18, 5 17, 6 16, 7 15, 8 14, 9 13, 10 12, 11 11, 12 10, 13 9, 14 8, 15 7, 16 6, 17 5, 18 4, 19 3,20 = 18 possibilities repeat the steps above, you have 18/400
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