Probability

[Kavya]

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???????

[bigmonkey31]

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

[sairamGmat]

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...

[November Rain]

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

[bigmonkey31]

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

Definitely the fastest way to do it!

[analyst218]

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

[Kavya]

Thanks so so much everyone

[bpgen]

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?

[firdaus117]

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?

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.

[GMATSUCKER]

Good one !!! But I'm not sure whether Stuart Kovinsky will like this as this might not be a real GMAT question !!