If x = [1,2,3] ; y = [4,5,6] and z = [7,8,9,10].
If numbers are selected at random; whats the probability that xyz will be EVEN?
Answer is 8/9..... how???????
Probability
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 62
- Joined: Sat Jan 02, 2010 1:11 am
- Thanked: 3 times
- GMAT Score:730
not sure if this is the quickest way.. but you basically need to count.
For each x, you have 1/3 chance of picking each number. For each y, you have a 1/3 chance, and z a 1/4 chance.
Now... we have 3 options with x. And recall that for a product of numbers to be even, you just need ONE of the variables to be even.
These are the combos to get even numbers.
x=1, y=5, z=8,10 P=1/3 x 1/3 x 1/2
x=1, y=4,6, z=7,8,9,10 P=1/3 x 2/3 x 1
x=2, y=4,5,6, z=7,8,9,10 p=1/3 x 1 x 1
x=3, y=5, z =8,10 P=1/3 x 1/3 x 1/2
x=3, y=4,6, z=7,8,9,10 P=1/3 x 2/3 x 1
Add them all up, and you get 16/18 or 8/9
For each x, you have 1/3 chance of picking each number. For each y, you have a 1/3 chance, and z a 1/4 chance.
Now... we have 3 options with x. And recall that for a product of numbers to be even, you just need ONE of the variables to be even.
These are the combos to get even numbers.
x=1, y=5, z=8,10 P=1/3 x 1/3 x 1/2
x=1, y=4,6, z=7,8,9,10 P=1/3 x 2/3 x 1
x=2, y=4,5,6, z=7,8,9,10 p=1/3 x 1 x 1
x=3, y=5, z =8,10 P=1/3 x 1/3 x 1/2
x=3, y=4,6, z=7,8,9,10 P=1/3 x 2/3 x 1
Add them all up, and you get 16/18 or 8/9
-
- Senior | Next Rank: 100 Posts
- Posts: 39
- Joined: Thu Feb 25, 2010 7:04 am
- Thanked: 2 times
This one is simple actually...we got to find out combinations of numbers with which product xyz will be odd
x = {1,2,3} y= {4,5,6} z= {7,8,9,10}
here the combinations with which we get odd number are:
x - y - z - product
1 - 5 - 7 - 35
1 - 5 - 9 - 45
3 - 5 - 9 - 135
3 - 5 - 7 - 105
So, there are 4 combinations with which we can get an odd number... As we know the probability of selecting a number from x, y and z are 1/3, 1/3 and 1/4 respectively...
The probability that we can have a odd number are ( 2/3) * (1/3) * (2/4) = 4/36 = 1/9.
So, the probability that we can have an even number is 1-1/9 = 8/9.
Why I wrote 2/3 for x => numbers which can result in ODD product are {1.3}
1/3 for y => numbers which can result in ODD product are {5}
2/4 for z => numbers which can result in ODD product are { 7,9}
That's it...
x = {1,2,3} y= {4,5,6} z= {7,8,9,10}
here the combinations with which we get odd number are:
x - y - z - product
1 - 5 - 7 - 35
1 - 5 - 9 - 45
3 - 5 - 9 - 135
3 - 5 - 7 - 105
So, there are 4 combinations with which we can get an odd number... As we know the probability of selecting a number from x, y and z are 1/3, 1/3 and 1/4 respectively...
The probability that we can have a odd number are ( 2/3) * (1/3) * (2/4) = 4/36 = 1/9.
So, the probability that we can have an even number is 1-1/9 = 8/9.
Why I wrote 2/3 for x => numbers which can result in ODD product are {1.3}
1/3 for y => numbers which can result in ODD product are {5}
2/4 for z => numbers which can result in ODD product are { 7,9}
That's it...
-
- Senior | Next Rank: 100 Posts
- Posts: 50
- Joined: Tue Oct 13, 2009 1:25 am
- Thanked: 14 times
- GMAT Score:720
Perhaps we coud do a different approach
Instead of computing the probability of an event occuring, perhaps we could do the probability of the opposite event.
For a number to be even in a multiplication, at least one factor must be even, so this means that for a number to be odd no factor can be even.
So, we have the following probabilities:
for X, there are 2/3 of probabilities of comming out odd
for Y, there are 1/3 of probabilities of comming out odd
for Z, there are 1/2 of probabilities of comming out odd
So the answer would be 1 - (2/3 * 1/3 * 1/2) , which is 8/9
Instead of computing the probability of an event occuring, perhaps we could do the probability of the opposite event.
For a number to be even in a multiplication, at least one factor must be even, so this means that for a number to be odd no factor can be even.
So, we have the following probabilities:
for X, there are 2/3 of probabilities of comming out odd
for Y, there are 1/3 of probabilities of comming out odd
for Z, there are 1/2 of probabilities of comming out odd
So the answer would be 1 - (2/3 * 1/3 * 1/2) , which is 8/9
-
- Senior | Next Rank: 100 Posts
- Posts: 62
- Joined: Sat Jan 02, 2010 1:11 am
- Thanked: 3 times
- GMAT Score:730
Definitely the fastest way to do it!November Rain wrote:Perhaps we coud do a different approach
Instead of computing the probability of an event occuring, perhaps we could do the probability of the opposite event.
For a number to be even in a multiplication, at least one factor must be even, so this means that for a number to be odd no factor can be even.
So, we have the following probabilities:
for X, there are 2/3 of probabilities of comming out odd
for Y, there are 1/3 of probabilities of comming out odd
for Z, there are 1/2 of probabilities of comming out odd
So the answer would be 1 - (2/3 * 1/3 * 1/2) , which is 8/9
-
- Master | Next Rank: 500 Posts
- Posts: 140
- Joined: Fri Feb 05, 2010 2:43 pm
- Thanked: 3 times
- GMAT Score:720
for sure.
1 - [all X Y Z odd] = even XYZ
1- [2/3 * 1/3 * 2/4] = 1- 1/9 = 8/9
number properties + simple probability
1 - [all X Y Z odd] = even XYZ
1- [2/3 * 1/3 * 2/4] = 1- 1/9 = 8/9
number properties + simple probability
- bpgen
- Master | Next Rank: 500 Posts
- Posts: 142
- Joined: Sat Feb 20, 2010 7:23 pm
- Thanked: 8 times
- Followed by:1 members
November Rain wrote:Perhaps we coud do a different approach
Instead of computing the probability of an event occuring, perhaps we could do the probability of the opposite event.
For a number to be even in a multiplication, at least one factor must be even, so this means that for a number to be odd no factor can be even.
So, we have the following probabilities:
for X, there are 2/3 of probabilities of comming out odd
for Y, there are 1/3 of probabilities of comming out odd
for Z, there are 1/2 of probabilities of comming out odd
So the answer would be 1 - (2/3 * 1/3 * 1/2) , which is 8/9
Hey November Rain,
Your posting looks interesting but could you please clarify my doubts below:
I think we could also count probabilities of even nos that coming out for each set as following:
for X, there are 1/3 of probabilities of coming out even
for Y, there are 2/3 of probabilities of coming out even
for Z, there are 2/4=>1/2 of probabilities of coming out even
then XYZ=1/9, do you know anything I'm missing out?
"Ambition is the path to success. Persistence is the vehicle you arrive in."
- firdaus117
- Master | Next Rank: 500 Posts
- Posts: 102
- Joined: Sat Feb 20, 2010 5:38 am
- Location: IIM Ahmedabad
- Thanked: 10 times
You have calculated the probability of xyz being even when all three digits are even.As a result you got lesser value than what it should be actually.Now,you know any single digit if even would turn product into even.That is why it is easier to calculate the opposite-when all three are odd,then only product is odd.bpgen wrote: Hey November Rain,
Your posting looks interesting but could you please clarify my doubts below:
I think we could also count probabilities of even nos that coming out for each set as following:
for X, there are 1/3 of probabilities of coming out even
for Y, there are 2/3 of probabilities of coming out even
for Z, there are 2/4=>1/2 of probabilities of coming out even
then XYZ=1/9, do you know anything I'm missing out?
-
- Senior | Next Rank: 100 Posts
- Posts: 46
- Joined: Sat Feb 27, 2010 2:58 am
- Location: GMAT
Good one !!! But I'm not sure whether Stuart Kovinsky will like this as this might not be a real GMAT question !!
GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT
What's life without GMAT !!!!!!!!
What's life without GMAT !!!!!!!!