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

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

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27 Aug 2012, 11:15
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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!
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Re: A three-digit positive integer is chosen at random.  [#permalink]

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27 Aug 2012, 11:58
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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

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

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27 Aug 2012, 12:37
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
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Re: A three-digit positive integer is chosen at random.  [#permalink]

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27 Aug 2012, 12:52
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]

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27 Aug 2012, 13:56
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]

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25 Sep 2012, 08:26
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]

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25 Sep 2012, 09:17
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]

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13 Sep 2016, 07:07
why it could not be 5C3 / 10C3 . I am failing to understand. If someone can please help.

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

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20 Apr 2018, 00:18
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Re: A three-digit positive integer is chosen at random. What is &nbs [#permalink] 20 Apr 2018, 00:18
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# A three-digit positive integer is chosen at random. What is

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