Difference between revisions of "2021 Fall AMC 10B Problems/Problem 14"

m (Solution)
(Tag: New redirect)
 
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 <math>4</math>. We have <math>2</math> 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 <math>6</math> 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>.
 
 
 
~kingofpineapplz
 
 
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=15|num-b=13}}
 
{{MAA Notice}}
 

Latest revision as of 00:31, 24 November 2021