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?
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 3 of 7.
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.
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)
Haridev Purve said:
7 years ago
Odd * Even = Even.
Total=(3+3+3)=9;
Even*odd=even
Total=(6+6+6)=18;
So here total odd number is (1,3,5);
And even (2,4,6);
Now E=27;
And Total=6*6=36
Prob=27/37=3/4 ans.
Total=(3+3+3)=9;
Even*odd=even
Total=(6+6+6)=18;
So here total odd number is (1,3,5);
And even (2,4,6);
Now E=27;
And Total=6*6=36
Prob=27/37=3/4 ans.
Naren said:
9 years ago
It may be also a short method:
Odd no 1st dice 1, 3, 5 = 3. 2nd dice odd no 1, 2, 3 = 3 hence 3 * 3 = 9.
Therefore, the product of even no 1-9/36 = 34.
Odd no 1st dice 1, 3, 5 = 3. 2nd dice odd no 1, 2, 3 = 3 hence 3 * 3 = 9.
Therefore, the product of even no 1-9/36 = 34.
Krishna said:
8 years ago
Sum of (4, 1), (4, 3), (4, 5) and so on in where ever even and add exit, the sum of two numbers becomes odd, but not even 4+1=5. So on with rest.
Isha said:
1 decade ago
1st dice * 2nd dice.
Even*even = even.
Even*odd = even.
Odd*even = even.
So, 3/6 * 3/6 + 3/6 * 3/6 + 3/6 * 3/6 = (3*3*3)/(6*6) = 3/4.
Even*even = even.
Even*odd = even.
Odd*even = even.
So, 3/6 * 3/6 + 3/6 * 3/6 + 3/6 * 3/6 = (3*3*3)/(6*6) = 3/4.
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.
I think this dice inside the value should multiply to get even values.
Eg: (1, 2) ---> (1×2) = 2 is an even number.
(3)
Shraddha pandey said:
1 decade ago
We can simply solve this problem by firstly finding the total outcomes and then possible events according to problem.
@pweety said:
1 decade ago
can it be solved like this: 6*6 = 36
0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36 = 18/36 = 9/12 = 3/4.
0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36 = 18/36 = 9/12 = 3/4.
Pravin atrekar said:
9 years ago
Wrong answer is mentioned here because we have to find the product not sum who's output comes even so and 9/35 =1/4.
Adi said:
9 years ago
Question is what are the chances of getting two numbers whose product (multiplication) is even number (2, 4, 6).
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