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

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

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?

thanks bunuel for your guidance

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 [X even x Y anything + Y even x X anything + both X and Y even]

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

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

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.


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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) = \(\frac{3}{4}\)

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.