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Sagot :
The two consecutive odd numbers that have a product of 15 are 3 and 5
There are 100 random 2 digits numbers [00 , 99]
There are 34 divisible by 3 {00, 03, 06, 09, … 93, 96, 99}
There are 20 divisible by 5 {00, 05, 10, … 90, 95}
However we must avoid counting numbers twice so we need to subtract those divisible by 15.
There are 7 divisible by 15 {00, 15, 30, 45, 60, 75, 90}
This means there are 34 + 20 - 7 = 47 2-digit numbers that are divisible by 3 or 5.
If you pick a random 2-digit number then P(divisible by 3 or 5) = 47/100 = 0.47
There are 100 random 2 digits numbers [00 , 99]
There are 34 divisible by 3 {00, 03, 06, 09, … 93, 96, 99}
There are 20 divisible by 5 {00, 05, 10, … 90, 95}
However we must avoid counting numbers twice so we need to subtract those divisible by 15.
There are 7 divisible by 15 {00, 15, 30, 45, 60, 75, 90}
This means there are 34 + 20 - 7 = 47 2-digit numbers that are divisible by 3 or 5.
If you pick a random 2-digit number then P(divisible by 3 or 5) = 47/100 = 0.47
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