If x is to be chosen at random from the set {1,2,3}, y is to be chosen

Question

If x is to be chosen at random from the set {1,2,3}, y is to be chosen at random from the set {4,5,6}, and z is to be chosen at random from the set {7,8,9,10}, what is the probability that xyz will be even?

A. 1/9

B. 1/2

C. 2/3

D. 7/9

E. 8/9

A. 1/9

B. 1/2

C. 2/3

D. 7/9

E. 8/9

QuantMadeEasy · Answer

Step 1: Understanding the question 
xyz is even when at-least one of them even
Probability of at-least one even = 1 - Probability of none even

Probability of none even = (Probability of x is odd) and (Probability of y is odd) and (Probability of z is odd)
 = 2/3 * 1/3 * 2/4 = 2/3 * 1/3 * 1/2 = 1/9

Probability of at-least one even = 1 - Probability of none even = 1 - 1/9 = 8/9

IMO E

ashimakumari · Answer

Why do we need to check whether x, y and z are even?
I think if xyz is a number then only units digit needs to be even, right?

Or did I misinterpret the question?

IanStewart · Answer

I'm guessing you're thinking that "xyz" is a three-digit number, where x is the hundreds digit, y the tens digit, and z the ones digit. That's not the intention here; here xyz is the product (multiplication) of x, y and z. Unless a question specifically tells you that letters represent digits in a number, then if you see something like cd or wxyz, it always means you're multiplying letters together.

And while I don't know if this will always be true (strictly speaking, it doesn't need to be), the only real GMAT questions I can recall where letters represent digits use capital letters. So if you see something like "ABC", then that  sometimes  (but only if the question also says so) will represent a three-digit number. But if the question doesn't tell you A, B and C are digits, then ABC would always mean the product A*B*C. I can't recall a single official question which uses lower-case letters ("abc") to represent digits, though I can't absolutely promise there aren't any. It is a convention in math to use capital letters for digits, but I'd imagine many prep company questions would not follow that convention.

varshas044 · Answer

If we directly jump into making the cases where x*y*z will be even --- there would be just too many possibilities to calculate.

Instead let us look at what cases to avoid

So we do not want x*y*z as odd, which is only possible when x, y and z all are odd.

X*y*Z (even) = 1 – (prob. Of x*y*z being odd)

prob. Of x*y*z being odd = Favorable / total 
total = 3c1 (choosing x) * 3c1 (choosing y) * 4c1 (choosing z) = 36 
favorable = 2c1 (x odd) * 1c1 (y odd) * 2c1 (z odd) = 4

X*y*Z (even) = 1 – (prob. Of x*y*z being odd) = 1 – 4/36 = 1-1/9 = 8/9

Answer = E

BrushMyQuant · Answer

Given that x is to be chosen at random from the set {1,2,3}, y is to be chosen at random from the set {4,5,6}, and z is to be chosen at random from the set {7,8,9,10} and we need to find what is the probability that xyz will be even

We know that Total Probability is always equal to 1

=>  P(Odd) + P(Even) = 1 
=>  P(Even) = 1 - P(Odd)

To get xyz as odd we need to have all of them as odd.
=> P(xyz = Odd) = P(x = odd) * P(y = odd) * P(z = odd) =  \frac{2}{3}  *  \frac{1}{3}  *  \frac{2}{4}

(Because for x we have two choices (1,3) to get odd out of 3, for y we have one choice (5) to get odd out of 3 and for z we have two choices (7,9) to get odd out of 4)

=  \frac{1}{9}

=>  P(Even) = 1 - P(Odd)  = 1 -  \frac{1}{9}  =  \frac{9 - 1}{9}  =  \frac{8}{9}

So,  Answer will be E .
Hope it helps!

RastogiSarthak99 · Answer

I did it slightly differently.

Total ways = 3 * 3 * 4 = 36 ways

1-P(even) = P(odd)

P(even) = (2c1 * 1c1 * 2c1)/36 = 1/9

1-1/9 = 8/9 = 88%

E

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

If x is to be chosen at random from the set {1,2,3}, y is to be chosen

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

If x is to be chosen at random from the set {1,2,3}, y is to be chosen

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards