|
|
| (2 intermediate revisions by the same user not shown) |
| Line 1: |
Line 1: |
| − | ==Problem 14==
| + | #REDIRECT [[2021_Fall_AMC_12B_Problems/Problem_11]] |
| − | Una rolls <math>6</math> standard <math>6</math>-sided dice simultaneously and calculates the product of the <math>6{ }</math> numbers obtained. What is the probability that the product is divisible by <math>4?</math>
| |
| − | | |
| − | <math>\textbf{(A)}\: \frac34\qquad\textbf{(B)} \: \frac{57}{64}\qquad\textbf{(C)} \: \frac{59}{64}\qquad\textbf{(D)} \: \frac{187}{192}\qquad\textbf{(E)} \: \frac{63}{64}</math>
| |
| − | | |
| − | ==Solution==
| |
| − | We will first find the probability that the product is not divisible by 4. We have 2 cases.
| |
| − | | |
| − | Case 1: The product is not divisible by <math>2</math>.
| |
| − | | |
| − | We need every number to be odd, and since the chance we roll an odd number is <math>\frac12,</math> our probability is <math>\left(\frac12\right)^6=\frac1{64}.</math>
| |
| − | | |
| − | | |
| − | | |
| − | Case 2: The product is divisible by <math>2</math>, but not by <math>4</math>.
| |
| − | | |
| − | We need <math>5</math> numbers to be odd, and one to be divisible by <math>2</math>, but not by <math>4</math>. There is a <math>\frac12</math> chance that an odd number is rolled, a <math>\frac13</math> chance that we roll a number satisfying the second condition (only <math>2</math> and <math>6</math> work), and 6 ways to choose the order in which the even number appears.
| |
| − | | |
| − | Our probability is <math>\left(\frac12\right)^5\left(\frac13\right)\cdot6=\frac1{16}.</math>
| |
| − | | |
| − | Therefore, the probability the product is not divisible by <math>4</math> is <math>\frac1{64}+\frac1{16}=\frac{5}{64}</math>.
| |
| − | | |
| − | Our answer is <math>1-\frac{5}{64}=\boxed{\textbf{(E)}\frac{59}{64}}</math>.
| |
| − | ==See Also==
| |
| − | {{AMC10 box|year=2021 Fall|ab=B|num-a=15|num-b=13}}
| |
| − | {{MAA Notice}}
| |
