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If $n$ positive integers taken at random are multiplied together, find the probability of the last digit of product being 5.

I know this question has been answered here. But I have tried it using a different approach and my answer doesn't match:

My approach:

Since there are $n$ positive integers, the number of ways their last digit can be chosen is ${10}^{n}$, so this becomes the sample space of the event, now for last digit of the product to be 5, I have chosen any one number from the sample space as $nC1$ and put its unit digit as 5, now for the remaining $n-1$ numbers the last digit can be $1,3,5,7,9$, this can be done in ${5}^{n-1}$ ways, making the total favourable cases to be $n{5}^{n-1}$, so the probability should be $$\frac{n{5}^{n-1}}{{10}^{n}}$$ But my answer does not match, can anyone please point out the error which I have made in my approach?

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Shriom707
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    You are counting each product with more than one five as many times as the number of fives in the product, depending on which of the fives you designate as the five in the product. – N. F. Taussig Aug 11 '20 at 08:58
  • I didn't understand what you wrote, please elaborate – Shriom707 Aug 11 '20 at 09:05
  • For $n=2$, the final digits of both numbers could be $5$. In your enumeration you count it twice: choosing the first number equal $5$ with the second number as one of ${1,3,5,7,9}$, and also as choosing the second number equal $5$ with the first number as one of ${1,3,5,7,9}$. – Jaap Scherphuis Aug 11 '20 at 09:16

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