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.
![]() |
n(E) | = | 27 | = | 3 | . |
n(S) | 36 | 4 |
Discussion:
67 comments Page 2 of 7.
Kranthi said:
1 decade ago
1 2 3 4 5 6
---------------------
1- 2 3 4 5 6 7
2- 3 4 5 6 7 8
3- 4 5 6 7 8 9
4- 5 6 7 8 9 10
5- 6 7 8 9 10 11
6- 7 8 9 10 11 12
using this table ec can do any sort of problems
e.g; we need sun 7 means check in table that how many times it comes
it is 6 there fore 6/36
---------------------
1- 2 3 4 5 6 7
2- 3 4 5 6 7 8
3- 4 5 6 7 8 9
4- 5 6 7 8 9 10
5- 6 7 8 9 10 11
6- 7 8 9 10 11 12
using this table ec can do any sort of problems
e.g; we need sun 7 means check in table that how many times it comes
it is 6 there fore 6/36
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
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
Vassu426 said:
1 decade ago
Dice contain 1,2,3,4,5,6 faces
from this 2,4,6 are even.
so product even is possible when,
even * even + odd * even + even * odd
so 1 has 3 chances of getting even i.e (1*2 ,1*4, 1*6)
2 has 6 chances
3 has 3
4 has 6
5 has 3
6 has 6
so total possible even products are= 3+6+3+6+3+6 => 27
ans : 27/36
=>3/4
from this 2,4,6 are even.
so product even is possible when,
even * even + odd * even + even * odd
so 1 has 3 chances of getting even i.e (1*2 ,1*4, 1*6)
2 has 6 chances
3 has 3
4 has 6
5 has 3
6 has 6
so total possible even products are= 3+6+3+6+3+6 => 27
ans : 27/36
=>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.
Ch. Anusha said:
1 decade ago
I do not understand because it is too lengthy process. Please tell the shortcuts.
Bharath. K said:
1 decade ago
Hi @Anusha,
I will try to make you understand the concept.
Probability is n(e)/n(S). here n(S) = 6*6 = 36.
To get n(e) i.e even product, we have to multiply odd with even or even with even.
Therefore, for odd with even we get,
1*2=2(product is even).
1*4.
1*6.
Therefore 3 combinations per odd number.
Similarly (do the same with 3 and 5) we get 3*3=9 combinations
for even with even: we get,
2*1.
2*2..till 2*6.
Therefore total 6 combinations with 3 even numbers, doing the same with 4 and 6 we get 6+6+6=18 combinations. Total n(e) =odd*even+even*even=9+18.
Substituting this value in n(e)/n(s) = 27/36 or 3/4.
I will try to make you understand the concept.
Probability is n(e)/n(S). here n(S) = 6*6 = 36.
To get n(e) i.e even product, we have to multiply odd with even or even with even.
Therefore, for odd with even we get,
1*2=2(product is even).
1*4.
1*6.
Therefore 3 combinations per odd number.
Similarly (do the same with 3 and 5) we get 3*3=9 combinations
for even with even: we get,
2*1.
2*2..till 2*6.
Therefore total 6 combinations with 3 even numbers, doing the same with 4 and 6 we get 6+6+6=18 combinations. Total n(e) =odd*even+even*even=9+18.
Substituting this value in n(e)/n(s) = 27/36 or 3/4.
Chiru said:
1 decade ago
Still it takes long time know?
I am getting concept but say in short way.
I am getting concept but say in short way.
Isha said:
1 decade ago
1st dice * 2nd dice.
Even*even = even.
Even*odd = even.
Odd*even = even.
So, 3/6 * 3/6 + 3/6 * 3/6 + 3/6 * 3/6 = (3*3*3)/(6*6) = 3/4.
Even*even = even.
Even*odd = even.
Odd*even = even.
So, 3/6 * 3/6 + 3/6 * 3/6 + 3/6 * 3/6 = (3*3*3)/(6*6) = 3/4.
Narendra said:
1 decade ago
Hi @bharath. K.
I am not understand below line.
n(S) = 6*6 = 36.
Why we are taking 6*6?
I am not understand below line.
n(S) = 6*6 = 36.
Why we are taking 6*6?
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