Aptitude - Probability - Discussion

Discussion Forum : Probability - General Questions (Q.No. 6)
6.
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
1
2
3
4
3
8
5
16
Answer: Option
Explanation:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
     (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
     (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) = n(E) = 27 = 3 .
n(S) 36 4

Discussion:
67 comments Page 1 of 7.

Kanishk said:   5 years ago
@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.
(51)

Ash said:   1 year ago
@All.

Here, (since odd * odd = odd).

Odd no's = 1, 3, 5.
1st dice P (not evn) = 3/6.
2nd dice P (not evn) = 3/6.
P (Not evn) = 3/6 * 3/6 = 1/4.
P (E) = 1-p (not E).
1-1/4 = 3/4.
(20)

ATUL PRAKASH said:   6 years ago
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.
(18)

Mohamed Hisham said:   2 years ago
Even no:2,4,6,8,10,12,14,.......

Two dice thrown of the product is:
= (1*2),(1*4)(1*6),(2*1),(2*2),(2*3),(2*4),(2*5),(2*6),(3*2),(3*4),(3*6)..........(6*6).
= 2,4,6,2,4,6,8,10,12,6,11,18,.......36.
n(E) = 27.
n(S) = 36.

So,
P(E) = n(E)/n(S).
= 27/36
= 3/4.
(12)

Soni kumari said:   3 years ago
I am not understanding this, please slove this in detail.
(5)

T.K. Dixit said:   6 years ago
The answer should be 1/2.
(4)

Saikiran kiran said:   5 years ago
I am not understanding this.
(4)

Habib rahuman said:   2 years ago
@All.

I think this dice inside the value should multiply to get even values.

Eg: (1, 2) ---> (1×2) = 2 is an even number.
(3)

Yebs said:   4 years ago
n(s) = 36*2=72.
n(E)=27*2.

Since it's two dice thrown simultaneously.
Then; 54/72 = 3/4.
(3)

Chetana said:   9 years ago
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.
(1)


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