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02 Nov 2021, 07:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
40% (03:04) correct
60%
(02:22)
wrong
based on 60
sessions
History
Date
Time
Result
Not Attempted Yet
A certain dice game can only be won if, when a player throws two fair six-sided dice, the number showing on one of the dice is a multiple of the number showing on the other. What is the probability that a player wins this game?
(A) 11/18
(B) 11/36
(C) 1/4
(D) 2/9
(E) 7/36
(A) 11/18
(B) 11/36
(C) 1/4
(D) 2/9
(E) 7/36
02 Nov 2021, 07:45
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You can notice that if you roll a '1' on the first die, the second number is always a multiple of your first roll, and if you roll anything else on the first die, there will always be at least two numbers you can roll on the second -- you can roll the same number again, or you can roll 1. So the answer clearly must be larger than 2/6, and only answer A is remotely plausible.
It's slightly tedious to get the answer without choices, but there are lots of ways to go about breaking down the cases. This might not be the fastest, but one way: we have 6*6 = 36 possible rolls of two dice, if we think about rolling them one at a time. In 11 of those, one of the rolls is a '1' (not 12, because we don't count 1, 1 twice), and then one roll is certainly a multiple of the other. In 5 more of the rolls, one is a '2' and the other is 2, 4 or 6 (again we don't count 2, 2 twice). In 3 more rolls, one roll is a '3' and the other is a '3' or a '6'. Then we also have three further possibilities, where we roll two 4's, two 5's or two 6's. Adding up all the possibilities, we can get two rolls where one is a multiple of the other in 11 + 5 + 3 + 3 ways, or 22 ways, so the answer is 22/36 = 11/18.
It's slightly tedious to get the answer without choices, but there are lots of ways to go about breaking down the cases. This might not be the fastest, but one way: we have 6*6 = 36 possible rolls of two dice, if we think about rolling them one at a time. In 11 of those, one of the rolls is a '1' (not 12, because we don't count 1, 1 twice), and then one roll is certainly a multiple of the other. In 5 more of the rolls, one is a '2' and the other is 2, 4 or 6 (again we don't count 2, 2 twice). In 3 more rolls, one roll is a '3' and the other is a '3' or a '6'. Then we also have three further possibilities, where we roll two 4's, two 5's or two 6's. Adding up all the possibilities, we can get two rolls where one is a multiple of the other in 11 + 5 + 3 + 3 ways, or 22 ways, so the answer is 22/36 = 11/18.
07 Sep 2023, 04:08
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For1: any number. Then (1/6*1)
For2: 1,2,4,6. Then (1/6*4/6)
For3: 1,3,6.Then (1/6*3/6)
For4: 1,2,4.Then (1/6*3/6)
For5: 1,5.Then(1/6*2/6)
For6:1,2,3,6.Then(1/6*4/6)
We have to use or probability and add above: ANSWER IS :22/36 or 11/18 (A)
Posted from my mobile device
For2: 1,2,4,6. Then (1/6*4/6)
For3: 1,3,6.Then (1/6*3/6)
For4: 1,2,4.Then (1/6*3/6)
For5: 1,5.Then(1/6*2/6)
For6:1,2,3,6.Then(1/6*4/6)
We have to use or probability and add above: ANSWER IS :22/36 or 11/18 (A)
Posted from my mobile device
