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 2 of 7.
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
Ram said:
1 decade ago
Another method:
The product will only by odd if the both dies are odd. P(1st Dice is odd) = 1/2 P(2nd Dice is odd) = 1/2.
1/2 * 1/2 = p both are odd hence product is odd as both rolls are independent events. P(odd product) = 1/4
Hence p(even product) = 1 - 1/4 = 3/4.
The product will only by odd if the both dies are odd. P(1st Dice is odd) = 1/2 P(2nd Dice is odd) = 1/2.
1/2 * 1/2 = p both are odd hence product is odd as both rolls are independent events. P(odd product) = 1/4
Hence p(even product) = 1 - 1/4 = 3/4.
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)
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.
Rohan said:
1 decade ago
P(odd) = 1/2
P(even) = 1/2
u will get an even number in three ways: odd*even
or even*odd or even * even
Hence the required probability =(1/2*1/2) +(1/2*1/2) + (1/2*1/2)
= 3*1/4
= 3/4
P(even) = 1/2
u will get an even number in three ways: odd*even
or even*odd or even * even
Hence the required probability =(1/2*1/2) +(1/2*1/2) + (1/2*1/2)
= 3*1/4
= 3/4
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.
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
Anonymous said:
1 decade ago
We can do this in the following way.
Getting an even product can be done in 3 ways
E*E=E;
E*O=E;
O*E=E;
(3C1/6C1)*(3C1/6C1)=9/36;
E*O=E;
(3C1/6C1)*(3C1/6C1)=9/36;
O*E=E;
(3C1/6C1)*(3C1/6C1)=9/36;
TOTALLY=
3*9/36= 3/4
Getting an even product can be done in 3 ways
E*E=E;
E*O=E;
O*E=E;
(3C1/6C1)*(3C1/6C1)=9/36;
E*O=E;
(3C1/6C1)*(3C1/6C1)=9/36;
O*E=E;
(3C1/6C1)*(3C1/6C1)=9/36;
TOTALLY=
3*9/36= 3/4
Dipaksinh said:
7 years ago
First, we have to select from any of the six sides then we have only three sides even or odd so 6c1*3c1.
Then we have 3 even and 3 odd for selection so 3c1*3c1 so final.
6c1 * 3c1+3c1 * 3c1/36.
Then we have 3 even and 3 odd for selection so 3c1*3c1 so final.
6c1 * 3c1+3c1 * 3c1/36.
Ashraf said:
9 years ago
Even = odd * even + even * even.
Odd = odd * odd oly
So in a dice odd = 1 3 5.
Another dice odd 135.
135
135 matrix e combination 9.
Odd possibility = 9/36.
Even possibility = 27/36.
Odd = odd * odd oly
So in a dice odd = 1 3 5.
Another dice odd 135.
135
135 matrix e combination 9.
Odd possibility = 9/36.
Even possibility = 27/36.
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