Aptitude - Probability - Discussion

Hi @Anusha,

I will try to make you understand the concept.

Probability is n(e)/n(S). here n(S) = 6*6 = 36.

To get n(e) i.e even product, we have to multiply odd with even or even with even.

Therefore, for odd with even we get,

1*2=2(product is even).
1*4.
1*6.

Therefore 3 combinations per odd number.

Similarly (do the same with 3 and 5) we get 3*3=9 combinations
for even with even: we get,

2*1.
2*2..till 2*6.

Therefore total 6 combinations with 3 even numbers, doing the same with 4 and 6 we get 6+6+6=18 combinations. Total n(e) =odd*even+even*even=9+18.

Substituting this value in n(e)/n(s) = 27/36 or 3/4.

The simple method is : we have 3 odd number(1, 3 ,5) and 3 even number(2, 4, 6);
We know that (even* even=even): = (3 even number * 3 even number=9));
(odd* even=even):= (3 odd number * 3 even number=9);
(even*odd=even):=(3 even number * 3 odd number=9);
Total even sum=9+9+9 = 27,
Total possiblity of two dice are = 6*6 = 36 (because each dice have 6 faces),
Therefore probablity=27/36 = 3/4.


Note: Here I have taken even*odd and odd*even because either of dice may contain even or odd;i.e: 1st dice contain 1 and second will contain 2 or 1st will contain 2 and 2nd will contain 1.

1 2 3 4 5 6
---------------------

1- 2 3 4 5 6 7

2- 3 4 5 6 7 8

3- 4 5 6 7 8 9

4- 5 6 7 8 9 10

5- 6 7 8 9 10 11

6- 7 8 9 10 11 12
using this table ec can do any sort of problems
e.g; we need sun 7 means check in table that how many times it comes
it is 6 there fore 6/36

We can do this question in two ways:

If we are looking at how many odd ways that better and easy to think :
1,3,5 are the odd digits in dice
to get odd both numbers should be odd
so the number of chances for the first number is 3,
and for second also 3.

Because 1*1,3*3 etc are also odd.
So favourable outcomes are 3*3=9.
prob=9/36 ->odd probability,
for prob(even)= 1-prob(odd).

@All.

I have a more elegant solution.

Probability of occurrence of an odd number in one die is 3/6 or 1/2. In two die it will be 1/2+ 1/2 = 1/4.

So 1/4 is the total probability of occurrence of an odd number in both dies. Therefore, the probability of occurrence of even number as product will be 1 - 1/4 (since the product of odd-even and even-even is always even)= 3/4.

n(E)=9.
Total numbers on two dices = 6*6 = 36.

Number of events when the number on two dices are odd and their sum is even =3.

Number of events when the number on two dices are even and their sum is even =3.

So the probability of the sum of the number being even=(3*3/36)=(9/36). Now each dice has six numbers. So (9*6/36)=(6/4)=(3/2).

So the probability is (3/2).

Just keep in the mind that odd is (e.g: 1, 3, 6) multiply with the even number i.e. 3 times each to make even.

And the even number is multiply with the all numbers (1, 2.6).

Solution:

1, 3, 5 (3 times each) i.e -3*3 = 9 times.

2, 4, 6 (6 times each) i.e -3*6 = 18 times.

So total favorable event-18+9 = 27.

Total no of event = 36.

So answer is = 27/36 = 3/4.

Dice contain 1,2,3,4,5,6 faces
from this 2,4,6 are even.
so product even is possible when,
even * even + odd * even + even * odd
so 1 has 3 chances of getting even i.e (1*2 ,1*4, 1*6)
2 has 6 chances
3 has 3
4 has 6
5 has 3
6 has 6

so total possible even products are= 3+6+3+6+3+6 => 27
ans : 27/36
=>3/4

Probability of getting product of two numbers.

Even E= {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3,) (5, 5), (6, 2), (6, 4), (6, 6)} =18.

Total = (6 * 6) = 36.

P (E) =N (e)/N(s) = 18/36 = 1/2.

I'm confused to get the answer. Can anyone help me?

You get an even number when:

Even x even.
Odd x even.
Even x odd.
Even numbers in each dice that may occur: 2, 4, 6 - Three possibilities,
Odd numbers in each dice: 1,3,5 -three possible conditions.

So even x even becomes 3 x 3 = 9.
Similarly for the other 2.
So, in total 9 + 9 + 9 = 27/36.