When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what i

Question

When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what is the probability that the product of x and y selected is even?

A. 1/4 
B. 1/2 
C. 3/4 
D. 4/5 
E. 9/10

* A solution will be posted in two day.

chetan2u · Answer

Hi all,
Whenever we see these type of Qs, we should look at the way the calculation of PROB is easier and lesser prone to errors..
it could be finding P straight or P as 1-P'...
here 1-P' is better..
what are the ways product is odd..
 when both numbers are odd

lets see the Q..
there are 10 digits, out of which 5 are odd and 5 are even..
choosing 2 out of 5 odd numbers=5*5, it is not given it is without replacement or in other words that x and y are different..
total ways of picking two numbers=10*10..

P'= 5*5/10*10=1/4..
so P, probability that the product of x and y selected is even=1-1/4=3/4
ans C

MathRevolution · Answer

When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what is the probability that the product of x and y selected is even?
 
 A. 1/4 B. 1/2 C. 3/4 D. 4/5 E. 9/10

-> Probability=the number of cases of a certain event/the number of the whole cases.
The probability of multiplication becoming a even number=1-the probability of multiplication becoming an odd number=1-(5C1*5C1/10C1*10C1)=1-(5*5/10*10)=1-1/4=3/4. Therefore, the answer is C.

rhine29388 · Answer

i have a doubt if we multiply any number from 1 to 9 with 0 the product rendered will be 0 are we considering this factor in our calculations?

Bunuel · Answer

ZERO: 1. 0 is an integer. 2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer except 0 itself. Check more here: number-properties-tips-and-hints-174996.html

rhine29388 · Answer

thanks bunuel for your guidance

deeuce · Answer

Hi  chetan2u   MathRevolution  :

I had a quick question on the approach for this problem. Here's how I calculated the solution:

for a product of X x Y to be even, either x has to be even or Y has to be even or both can be even.:

so I took 5C1/10C1 x 10C1/10C1 + 5C1/10C1 x 10C1/10C1 + 5C1 x 5C1 / 10C1 x 10C1

=> 1/2 x 1 + 1/2 x 1 + 1/4 => 5/4

Can you help me identify where I'm going wrong with this?

chetan2u · Answer

Hi

When you take x as even and y as anything, the total includes both even.
Similarly, when you take y as even and x as anything, the total includes both even.

Thus, in above scenarios,  'both even'  is getting added twice, so you require to subtract that once.
However, you have added that once more.

vaibhav1221 · Answer

There are two ways to go about such questions. The fool proof way is to subtract the unfavourable probability from 1.

Method 1: Listing and adding favourable cases. 
There are four possibilities
 X = Even; Y = Odd -----> XY = Even
X = Odd; Y = Even -----> XY = Even
X = Even; Y = Even -----> XY = Even 
 X = Odd; Y = Odd -----> XY = Odd

3 out of the 4 cases are favourable. Hence, the probability is 3/4.

Alternatively, You could also try to do a little math and solve the following way: 
P(x=even) =  \frac{1}{2}  = P(x=odd)
P(y=odd) =  \frac{1}{2}  = P(y=even)
P(xy = even) = P(x=odd; y=even) + P(x=even; y=odd) + P(x=even; y=even)
P(xy = even) =  \frac{1}{2}*\frac{1}{2} + \frac{1}{2}*\frac{1}{2} + \frac{1}{2}*\frac{1}{2} 
P(xy = even) =  \frac{1}{4} + \frac{1}{4} + \frac{1}{4} 
P(xy = even) =  \frac{3}{4}

Method 2: Subtracting the unfavourable probability from 1. 
P(xy = odd) = P(x=odd; y=odd)
P(xy = odd) =  \frac{1}{2*1/2} 
P(xy = odd) =  \frac{1}{4}

P(xy=even) = 1 - P(xy=odd)
P(xy=even) =  1 - \frac{1}{4} 
P(xy=even) =  \frac{3}{4}

charlie07 · Answer

x can be selected in 10 ways, and so can y be.
Over all there are 100 ways of choosing x and y.

Probability of x*y to be even = 1- Probability of x*y to be odd

finding Probability of x*y to be odd is easier. Both x and y have to be odd.
odd x can be selected in 5 ways (1,3,5,7,9). So can odd y be.
So, Probability of x*y to be odd = 5*5/10*10 = 1/4

Probability of x*y to be even = 1-1/4 = 3/4

so, C is correct answer.

Vivek1707 · Answer

Hi  chetan2u

In this case its not mentioned that replacements are allowed. So is it safe to assume that replacements are allowed whenever its not mentioned ? . Please guide

Kinshook · Answer

When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what is the probability that the product of x and y selected is even?

The probability that the product of x & y selected is even = 1 - The probability that the product of x & y selected is odd = 1 - 5*5/(10*10) = 1 - 1/4 = 3/4

IMO C

vasu1104 · Answer

set = 0,1,2,3,4,5,6,7,8,9

even=0,2,4,6,8
odd= 1,3,5,7,9

there are 2 ways to approach here.

start with less time consuming one.
find probability of finding odd product.
for that both x and y need to be odd.

1- (5/10)(5/10)
3/4

approach 2

in this there are 3 cases.
case 1=
x= even, y= even
5/10 * 5/10 = 1/4

case 2=
x- even, y- odd
5/10 * 5/10 = 1/4

case 3=
x-odd, y-even
5/10 * 5/10 =1/4

add all 3 cases together.
1/4+1/4+1/4
3/4

choice C

When x, y are selected randomly from set {0,1,2,3,4,5,6,7,8,9}, what i

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