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 4 of 7.
Swathy said:
9 years ago
Really useful information for non-maths students like me. Thank to all.
Prateek said:
9 years ago
I have the same question. Just like Keisha. Please, someone give answer.
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.
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)
Keisha said:
9 years ago
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?
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?
Adi said:
9 years ago
Question is what are the chances of getting two numbers whose product (multiplication) is even number (2, 4, 6).
Tarunaa said:
9 years ago
I am unable to understand.
Please someone explain in briefly!
Please someone explain in briefly!
Sabyasachi said:
10 years ago
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.
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.
Shubhankar said:
10 years ago
Simply do.
1 = 3 times even nos (ex-1*2=2, 1*4=4, 1*6=6).
2 = 6 times.
3 = 3 times.
4 = 6 times (4*1=4, 4*2=8. Etc).
5 = 3 times even nos.
6 = 6 times.
Total times even numbers = 27; n(s) = 6*6 = 36.
Now by formula p(e) = N(E)/N(S).
27/36 = 3/4.
1 = 3 times even nos (ex-1*2=2, 1*4=4, 1*6=6).
2 = 6 times.
3 = 3 times.
4 = 6 times (4*1=4, 4*2=8. Etc).
5 = 3 times even nos.
6 = 6 times.
Total times even numbers = 27; n(s) = 6*6 = 36.
Now by formula p(e) = N(E)/N(S).
27/36 = 3/4.
Ziku said:
1 decade ago
Product must be a number which might be an even or an odd. So P(even/odd) = 1/2.
NIKHIL said:
1 decade ago
Sir please any one explain in a question ask product of even number but in solution shown odd number.
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