Difference between revisions of "2019 AIME II Problems/Problem 4"
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(1/3)^4 * 1 + 4 * (1/3)^3 * (2/3) * 0 + 6 * (1/3)^2 * (2/3)^2 * 1/4 + 4 * (1/3) * (2/3)^3 * 3/32 + (2/3)^4 * 5/32 = 25/162. | (1/3)^4 * 1 + 4 * (1/3)^3 * (2/3) * 0 + 6 * (1/3)^2 * (2/3)^2 * 1/4 + 4 * (1/3) * (2/3)^3 * 3/32 + (2/3)^4 * 5/32 = 25/162. | ||
− | 25+162 = \boxed{187} | + | 25+162 = <cmath> \boxed{187} </cmath> |
-dnaidu (silverlizard) | -dnaidu (silverlizard) |
Revision as of 22:30, 22 March 2019
Contents
Problem 4
A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is , where and are relatively prime positive integers. Find .
Solution
Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are outcomes).
Case 1 (easy): Four 5's are rolled. This has probability of occurring.
Case 2: Two 5's are rolled.
Case 3: No 5's are rolled.
To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For , let equal the number of outcomes after rolling the die times, with the property that the product is a square. Thus, as 1 and 4 are the only possibilities.
To find given (where ), we observe that if the first rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives outcomes. Otherwise, the first rolls do not multiply to a perfect square ( outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first rolls is where and are not both even, then we observe that if and are both odd, then the last roll must be 6; if only is odd, the last roll must be 2, and if only is odd, the last roll must be 3. Thus, we have outcomes in this case, and .
Computing , , gives , , and . Thus for Case 3, there are 157 outcomes. For case 2, we multiply by to distribute the two 5's among four rolls. Thus the probability is
-scrabbler94
Solution 2
We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Lets call rolling 1 or 4 rolling a dud.
Probability of rolling 4 duds: (1/3)^4
Probability of rolling 3 duds: 4 * (1/3)^3 * (2/3)
Probability of rolling 2 duds: 6 * (1/3)^2 * (2/3)^2
Probability of rolling 1 dud: 4 * (1/3) * (2/3)^3
Probability of rolling 0 duds: (2/3)^4
Now we will find the probability of a square product given we have rolled a each amount of duds
Probability of getting a square product given 4 duds: 1
Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)
Probability of getting a square product given 2 duds: 1/4 (as long as our two non-duds are the same, our product will be square)
Probability of getting a square product given 1 duds: (3!/4^3) = 3/32 (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of 4^3 ways to roll 3 non-duds).
Probability of getting a square product given 0 duds: 40/4^4 = 5/32 (We can have any two non-duds twice. For example, 2,2,5,5. There are 4c2 ways of choosing which two non-duds to use and 4c2 ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).
We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values.
(1/3)^4 * 1 + 4 * (1/3)^3 * (2/3) * 0 + 6 * (1/3)^2 * (2/3)^2 * 1/4 + 4 * (1/3) * (2/3)^3 * 3/32 + (2/3)^4 * 5/32 = 25/162. 25+162 =
-dnaidu (silverlizard)
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.