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 1 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)
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)
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.
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)
Chetna Chauhan said:
3 years ago
Yes, It is easy for understanding. Thanks, everyone for explaining.
(1)
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.
n(E)=27*2.
Since it's two dice thrown simultaneously.
Then; 54/72 = 3/4.
(3)
Poulomi sen said:
5 years ago
Thanks for explaining the solution @Kanishk.
Raji Rajendran said:
5 years ago
@Kanishk.
It's understandable.
It's understandable.
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.
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)
Saikiran kiran said:
5 years ago
I am not understanding this.
(4)
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