Any a three-digit positive integer has AT LEAST one even number. In this sense, any number that doesn't have an even digit is odd.
Therea are 5 odds digits: 1,3, 5, 7, and 9.
So, the total number of numbers whose product is odd (all digits are odd) is: 5*5*5 =125
Also, the total number of numbers with three digits is: 9*10*10 = 900. (The hundreds digits only has 9 possibilities. It cannoy be 0).
Ok, we have everything, let's calculate:
\(\frac{(9*10*10 - 5*5*5)}{(9*10*10)} = 1 - \frac{(5*5*5)}{(9*10*10)} = 1 - \frac{5}{36} = \frac{31}{36}\)

Answer B.
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Here is a direct probabilistic approach, first using the probability of the complementary event. Denote the three-digit number by ABC, where A>0.

P(A*B*C even) = 1 - P(A*B*C odd) = 1 - P(A odd)*P(B odd)*P(C odd) = 1 - (5/9)*(5/10)*(5/10) = 1 - 5/36 = 31/36.

Now let's do it the long way, directly computing P(A*B*C even) = P((A even) OR (B even) OR (C even)).
We have to apply the formula for the union of three events:

\(P(A\cup{B}\cup{C})=P(A)+P(B)+P(C)-P(A\cap{B})-P(A\cap{C})-P(B\cap{C})+P(A\cap{B}\cap{C})\)

In our case, the above formula will lead to \(\frac{4}{9}+\frac{1}{2}+\frac{1}{2}-\frac{4}{9}\cdot\frac{1}{2}-\frac{4}{9}\cdot\frac{1}{2}-\frac{1}{2}\cdot\frac{1}{2}+\frac{4}{9}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{31}{36}.\)

Answer B
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The question asks us to calculate the probability of selecting a three digit number with digits whose product is even ..

We know that three digit numbers begin from 100 , and end with 999 ie. 900 total numbers...

Let us find all the numbers - the product of whose digits is odd .. This is only possible when all three digits are ODD ...

From 0 - 9 we have 1,3,5,7,9 as odd digits that is 5 odd digits

XXX , can have 5 x 5 x 5 possible values (the question speaks nothing of NOT REPEATING numbers) = 125 different numbers with the product of the digits being ODD ....

Now the Probability of finding odd + Probability of finding EVEN = 1

Probability of finding EVEN = 1 - Probability of finding ODD

We know that odd numbers can have 125 different combinations , and total combinations are 900 , therefore probability of finding odd = 125/900

Therefore Probability of finding even = 1 - (125/900) = 1 - 5/36 = 31/36 (B)
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why it could not be 5C3 / 10C3 . I am failing to understand. If someone can please help.

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Megha