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A machine randomly generates one of the nine numbers 1, 2, 3, … , 9 with equal likelihood. What is the probability that when Tsuni uses this machine to generate four numbers their product is divisible by 4? Express your answer as a common fraction.

 Jul 22, 2022
 #1
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Note that for the product to be divisible by 4, there need to be at least 2 even numbers. 


The probability of rolling only odd numbers is \({5 \over 9}^4 = {1025 \over 6561}\)

 

The probability of rolling exactly 1 even number is \(4 \times 5^3 \times {4 \choose 3} = 2000\)

 

So, the probability is \(1 - ({1025 \over 6561} + {2000 \over 6561}) = \text{_____}\)

 Jul 22, 2022
 #4
avatar+2339 
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Disregard this answer.....

BuilderBoi  Jul 23, 2022
 #2
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BuilderBoi: A computer code gives the following numbers:

 

9^4 ==6,561 permutations. [This is based on the assumption that numbers are replaced at each draw]

 

The computer calculates:

 

The probability is;  4,936 / 6,561 

 Jul 22, 2022
 #3
avatar+2339 
+1

REVISED ANSWER: There are \(9^4 = 6561\) ways to get the numbers. 

 

The easiest way to solve this is to use complementary counting, in which we count what we don't want, and subtract that from 1

 

Note that the probability of getting an odd number is \({5 \over 9}^4 = {625 \over 6561}\)

 

The probability of getting a 2 or 6 is \({2 \times 5^3 \times {4 \choose 3} \over 6561} = {1000 \over 6561}\) note that 4 and 8 make the product divisible by 4. 

 

So, the total probability is \(1 - ({625 \over 6561} + {1000 \over 6561} ) = \color{brown}\boxed{4936 \over 6561}\)

 Jul 23, 2022

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