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.
@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.
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.
Dhruvil said:
1 decade ago
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).
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).
Arijit said:
1 decade ago
Two odd number makes the product odd. i.e. (1,3,5).
Probability of being the product odd is (1/2)*(1/2) = 1/4.
As we have to see only probability in getting odd numbers for 2 dices.
Probability of getting the product even is 1-(1/4) = 3/4.
Probability of being the product odd is (1/2)*(1/2) = 1/4.
As we have to see only probability in getting odd numbers for 2 dices.
Probability of getting the product even is 1-(1/4) = 3/4.
Suman said:
1 decade ago
There is a formula 7-n & 13-n. For what this formula is used?
Avhi said:
1 decade ago
Why I'm choosing 6 number we have a another number?
Soumya said:
1 decade ago
There is not any either easy solution? Which must not be confusing?
NIKHIL said:
1 decade ago
Sir please any one explain in a question ask product of even number but in solution shown odd number.
Ziku said:
1 decade ago
Product must be a number which might be an even or an odd. So P(even/odd) = 1/2.
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.
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