A three-digit positive integer is chosen at random. What is

Question

A three-digit positive integer is chosen at random. What is the probability that the product of its digits is even?

A. 1/2
B. 31/36
C. 49/54
D. 7/8
E. 11/12

Hello,

I got this question from Thursdays with Ron. I wonder if someone could explain how I get the solution without the "1- not desired outcome" trick? What do you think is the problem's difficulty level?

Thanks!

metallicafan · Accepted Answer

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.

rocketscience · Answer

Thanks for your reply! Ron explained this way to solve the question, but what about:

ODD * ODD * EVEN + ODD * EVEN * EVEN + EVEN * EVEN * EVEN = (5 * 5 * 5 + 5 * 5 * 5 + 4 * 5 * 5)/900

This doesn't work, but I could not figure out where I am wrong!

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EvaJager · Answer

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

EvaJager · Answer

You have many more possibilities. You missed:
ODD*EVEN*ODD
EVEN*ODD*ODD

EVEN*ODD*EVEN
EVEN*EVEN*ODD

You should have 7 possibilities: 3 when there is one even digit, 3 when there are two even digits, and 1 when all three digits are even.

vomhorizon · Answer

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)

EvaJager · Answer

Why complicate it? Just compute the probability of the complementary event:

P(three odd digits) = 5/9*5/10*5/10 = 5/36 (first digit cannot be 0, so we have a total of 9 digits, 4 even and 5 odd).

Therefore, P(at least one even digit) = 1 - 5/36 = 31/36.

Answer B.

megha_2709 · Answer

why it could not be 5C3 / 10C3 . I am failing to understand. If someone can please help.

Regards Megha

BrushMyQuant · Answer

Total Even numbers can be found by taking even digit in any of the three places. So there will be lot of combinations and it will be difficult to list down everything.
This is why an approach to find odd number is easier and will save you time.
If you want to find total number of even numbers then they this.
First digit even (4 ways) * anything in second and third. So 400 ways.
Second digit even and first digit odd and anything in third place.
So, 5*5*10 = 250 ways
Third digit even and first and second are odd = 5*5*5 = 125 ways
Total = 400+250+125 = 775.

Hope it helps!

ThatDudeKnows · Answer

We "lose" only with OddOddOdd.
The probability of getting odd with the first digit is 5/9. Why 5/9? We can choose from 1-9 and five of those are odd.
The probability of getting odd with the second digit is 1/2. Why 1/2? We can choose from 0-9 and five of those are odd.
The probability of getting odd with the third digit is 1/2. Same.
The probability of "losing" is  \frac{5}{9}*\frac{1}{2}*\frac{1}{2}=\frac{5}{36} 
The probability of "winning" is  \frac{31}{36}

Answer choice B.

gmatpro15 · Answer

here instead of finding even numbers use complement rule

find numbers which are odd

we can get product as odd only if all the 3 numbers are odd 
so total odd digit are 1,3,5,7,9
since odd * odd * odd = odd
so getting odd = 5/9 x 1/2 X 1/2 = 5/36 (total odd numbers = 5, possible numbers = 9 (since we cant first digit as 0 since it wont be 3 digit number )

SO GETTING EVEN = 1- P(GETTING ODD) = 1-(5/36) = 31/36

LeelooMats · Answer

Hi,

I did it in the following way:

EEE = 1/2*1/2*1/2 = 1/8
EEO = 1/2*1/2*1/2 = 1/8* 3!/2! = 3/8
OOE = 1/2*1/2*1/2 = 1/8* 3!/2! = 3/8

Adding them = 7/8

What's my error? Thank You.

Krunaal · Answer

You're considering cases where first digit can be 0.

The best way to go about this problem is to find count of 3-digit numbers having ALL ODD digits, and subtracting them from Total 3-digit numbers to get count of 3-digit numbers whose product of digits is even.

PSKhore · Answer

There are 900 three-digit numbers. 
Multiplication of all odd numbers will lead to an odd number.
Therefore, OOO (1,3,5,7,9) = 5*5*5 = 125
(900-125)/900 
= 775/900 
= 31/36

tcadiz · Answer

For the product of the digits to be even, it only needs one digit that is even (if you factorize by 2 then it would be in the form 2*X*Y*Z). Therefore, the only odd results will be when all of the digits are odd. we only have 5 odd numbers to choose from each digit. 5*5*5 = 125. 1 - 125/900 = 31/36

Rickooreo · Answer

Build this trigger
Whenever you see:
 "at least one" / "product is even" 
immediately ask:
 "What would make this NOT happen?" 
Here, NOT even product =  all digits odd .
 Important:  Directly doing “hundreds even + tens even + units even” causes  double counting .

### Method 1: Complement Method — Faster

Product is even = At least one digit is even.

Opposite: Product is odd = ALL digits are odd.

Total 3-digit numbers = 900

Odd choices:
Hundreds digit = 5 choices (1, 3, 5, 7, 9)
Tens digit = 5 choices
Units digit = 5 choices

Numbers with odd product = 5 × 5 × 5 = 125

P(odd product) = 125/900 = 5/36

P(even product) = 1 − 5/36 = 31/36

Answer: B. 31/36

Important: When you see "at least one" or "product is even," think COMPLEMENT → "all are odd."

### Method 2: Direct Counting — First Even Digit

Product is even = At least one digit is even.

To avoid double counting, classify by where the FIRST even digit appears.

Case 1: Hundreds digit is even
= 4 × 10 × 10
= 400

Case 2: Hundreds odd, tens even
= 5 × 5 × 10
= 250

Case 3: Hundreds odd, tens odd, units even
= 5 × 5 × 5
= 125

Total favorable = 400 + 250 + 125 = 775

P(even product) = 775/900 = 31/36

Answer: B. 31/36

Important: Don't simply add "hundreds even + tens even + units even" because numbers such as 246 would be counted multiple times.

ZeroSquare · Answer

I got the answer straight forward by subtracting total - case in which all 3 numbers are odd in the integer. As that is the only case that would give a non even/odd product.

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