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What is the probability that a randomly selected integer in the set { 1, 2, 3 ... 100 } is divisible by 2 and not divisible by 6? Express your answer as a common fraction.

 Feb 4, 2022
 #1
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ugh...

 

When I wanted to click "Show Preview", they deleted my answer...

 

So... here we go again:

 

\(\frac{\mbox{Total number of all possible solutions}}{\mbox{Total number of all solutions}}\)

 

In the set, we can choose 100 numbers \(\Rightarrow\)\(\frac{n}{100}\), where n is a positive integer.

 

From those 100 numbers, ( 100 / 2 = ) 50 numbers are divisible by 2.

 

From those 50 numbers,

 

50 / 6 = 8 mod 2 \(\Rightarrow\) 8 numbers

 

are divisible by 6.

 

Subtract both numbers \(\Rightarrow\) 50 - 8 = 42 numbers, so

 

42 / 100 = 21 / 50 .

 

Thus, our answer is 21 / 50.

 Feb 4, 2022
 #2
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2 , 4 , 8 , 10 , 14 , 16 , 20 , 22 , 26 , 28 , 32 , 34 , 38 , 40 , 44 , 46 , 50 , 52 , 56 , 58 , 62 , 64 , 68 , 70 , 74 , 76 , 80 , 82 , 86 , 88 , 92 , 94 , 98 , 100>>Total==34 such integers.

 

The probability is: 34 / 100==17 / 50

 Feb 4, 2022

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