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Of all the four-digit positive integers containing only digits from the set {2,4,6,8} what fraction of them have at least one 8?

 May 22, 2021
 #1
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3 * 3 = 9

 

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We choose which diget is 4.

 May 22, 2021
 #2
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The total number of 4 digit numbers with digits from the set given is \((4)(4)(4)(4) ={4}^{4}.\) If we leave out 8, the total number changes to

\((3)(3)(3)(3)={3}^{4}\). So the fraction having at least one 8, which is 1 minus the fraction having no 8, would then be

\(1-\frac{{3}^{4}}{{4}^{4}}=1-\frac{81}{256}=\frac{175}{256}\)

 May 23, 2021

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