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Re: A three-digit positive integer is chosen at random. [#permalink]
rocketscience wrote:
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!

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


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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Re: A three-digit positive integer is chosen at random. [#permalink]
rocketscience wrote:
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!

Posted from GMAT ToolKit


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.
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Re: A three-digit positive integer is chosen at random. What is [#permalink]
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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Re: A three-digit positive integer is chosen at random. [#permalink]
EvaJager wrote:
rocketscience wrote:
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!

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


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


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.
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Re: A three-digit positive integer is chosen at random. What is [#permalink]
why it could not be 5C3 / 10C3 . I am failing to understand. If someone can please help. :(

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Re: A three-digit positive integer is chosen at random. What is [#permalink]
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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!
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Re: A three-digit positive integer is chosen at random. What is [#permalink]
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rocketscience wrote:
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!


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.
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Re: A three-digit positive integer is chosen at random. What is [#permalink]
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
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Re: A three-digit positive integer is chosen at random. What is [#permalink]
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Re: A three-digit positive integer is chosen at random. What is [#permalink]
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