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 7 of 7.
Phani said:
1 decade ago
Provide shortcuts also with these explanations then we can able to do solve the problems very fastly.
Vara said:
1 decade ago
For sum of numbers it is 18, for product of numbers its correct.
Thangamuthu said:
1 decade ago
How it is.....? only 18 even numbers combinations are possible ....?so ans must be 1/2. .
Jayanna said:
1 decade ago
This is the best method sir. the first method is confusion it will take time also . i want give u one suggestion also sir for this prob. there are 2,4,6 are even numbers in the die. for 2 ,there is 1,for 4, there is 3, and for 6 there is 5. sum of 1,3,6 equal to 9 .so 9/36=1/4 then 1-1/4=3/4
Sunny said:
1 decade ago
Ya srinivas here it is
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, not getting even no is e={(1,3)(1,5)(1,1)(3,1)(3,3)(3,5)(5,1)(5,3)(5,5)}
n(E)=9
p(E)=9/36=1/4
now probability of getting even is =1-(1/4)=3/4
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, not getting even no is e={(1,3)(1,5)(1,1)(3,1)(3,3)(3,5)(5,1)(5,3)(5,5)}
n(E)=9
p(E)=9/36=1/4
now probability of getting even is =1-(1/4)=3/4
Ranjan said:
1 decade ago
Please discuss more about it
Srinivas said:
2 decades ago
Is there any method for to solve all these type of questions?
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