3. A fair decahedral die has faces numbered 0-9, each equally likely to occur wh
ID: 3064492 • Letter: 3
Question
3. A fair decahedral die has faces numbered 0-9, each equally likely to occur when a. If two fair decahedral dice are rolled, what is the probability that the sum of the b. If three fair decahedral dice are rolled, what is the probability that the sum of the c. If eight fair decahedral dice are rolled, what is the probability that at least one of d. If a fair decahedral die is rolled 5 times, what is the probability that all the rolls e. If a fair decahedral die is rolled 7 times, what is the probability that only the last roll the die is rolled. Round each answer to 4 places after the decimal point. numbers rolled is at most 4? numbers rolled is 16? the numbers rolled is 2? result in a different number? is at least 6?Explanation / Answer
a.
The sum of the number of rolled is atmost 4 is when the outcomes of (Dice1, Dice2) are,
(0, 0)
(0, 1), (1, 0)
(0, 2), (1, 1) (2, 0)
(0, 3), (1, 2), (2, 1) (3, 0)
(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)
So, there are 15 outcomes out of 100 outcomes.
Probability that sum of the number of rolled is atmost 4 = 15 / 100 = 0.15
b.
The sum of the number of rolled is 16 is when the outcomes of (Dice1, Dice2, Dice3) are,
(0, 7, 9), (0, 8, 8), (0, 9, 7)
(1, 6, 9), (1, 7, 8), (1, 8, 7), (1, 9, 6)
(2, 5, 9), (2, 6, 8), (2, 7, 7), (2, 8, 6), (2, 9, 5)
(3, 4, 9), (3, 5, 8), (3, 6, 7), (3, 7, 6), (3, 8, 5), (3, 9, 4)
(4, 3, 9), (4, 4, 8), (4, 5, 7), (4, 6, 6), (4, 7, 5), (4, 8, 4), (4, 9, 3)
(5, 2, 9), (5, 3, 8), (5, 4, 7), (5, 5, 6), (5, 6, 5), (5, 7, 4), (5, 8, 3), (5, 9, 2)
(6, 1, 9), (6, 2, 8), (6, 3, 7), (6, 4, 6), (6, 5, 5), (6, 6, 4), (6, 7, 3), (6, 8, 2), (6, 9, 1)
(7, 0, 9), (7, 1, 8), (7, 2, 7), (7, 3, 6), (7, 4, 5), (7, 5, 4), (7, 6, 3), (7, 7, 2), (7, 8, 1), (7, 9, 0)
(8, 0, 8), (8, 1, 7), (8, 2, 6), (8, 3, 5), (8, 4, 4), (8, 5, 3), (8, 6, 2), (8, 7, 1), (8, 8, 0)
(9, 0, 7), (9, 1, 6), (9, 2, 5), (9, 3, 4), (9, 4, 3), (9, 5, 2), (9, 6, 1), (9, 7, 0)
So, there are 69 outcomes out of 1000 outcomes.
Probability that sum of the number of rolled is 16 = 69 / 1000 = 0.069
c.
Probability that 2 is rolled on any one dice = 1/ 10 = 0.1
Probability that 2 is not rolled on any one dice = 1 - 0.1 = 0.9
Probability that 2 is rolled atleast on any one of 8 dices = 1 - Probability that 2 is not rolled on any one of 8 dices
= 1 - 0.98 = 0.5695
d.
Number of ways to select 5 different numbers from 10 numbers = 10P5 = 10! / 5! = 30240
Number of possible outcomes from 5 rolls = 105 = 100000
Probability that all rolls result in a different number = 30240 / 100000 = 0.3024
e.
Probability that the roll of the dices is atleast 6 = 4/10 = 0.4 (Outcome is 6, 7, 8 or 9)
Probability that the roll of the dices is not atleast 6 = 1 - 0.4 = 0.6
Probability that only the last roll is atlest 6 = Probability that 1st 6 rolls is not atlest 6 * Probability that the last roll is atlest 6
= 0.66 * 0.4 = 0.0187
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