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.
Swathy said:
9 years ago
Really useful information for non-maths students like me. Thank to all.
Keisha said:
9 years ago
Probability of getting product of two numbers.
Even E= {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3,) (5, 5), (6, 2), (6, 4), (6, 6)} =18.
Total = (6 * 6) = 36.
P (E) =N (e)/N(s) = 18/36 = 1/2.
I'm confused to get the answer. Can anyone help me?
Even E= {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3,) (5, 5), (6, 2), (6, 4), (6, 6)} =18.
Total = (6 * 6) = 36.
P (E) =N (e)/N(s) = 18/36 = 1/2.
I'm confused to get the answer. Can anyone help me?
Panging said:
8 years ago
How n(E) got =27?
Krishna said:
8 years ago
Sum of (4, 1), (4, 3), (4, 5) and so on in where ever even and add exit, the sum of two numbers becomes odd, but not even 4+1=5. So on with rest.
Naveena said:
8 years ago
I can't understand it please explain this problem.
Zaheer said:
8 years ago
Is there any shortcut method to solve such types of question.
Anusuya said:
8 years ago
I don't understand why take product why only take sum?
Please, guys explain.
Please, guys explain.
Shishirkumar said:
8 years ago
3 cases.
even* odd=even.
even * even =even.
odd * even= even.
add all
(3/6*3/6)+3/6*3/6)+(3/6*3/6)=3/4.
even* odd=even.
even * even =even.
odd * even= even.
add all
(3/6*3/6)+3/6*3/6)+(3/6*3/6)=3/4.
Anjan vikas reddy said:
7 years ago
We can do this question in two ways:
If we are looking at how many odd ways that better and easy to think :
1,3,5 are the odd digits in dice
to get odd both numbers should be odd
so the number of chances for the first number is 3,
and for second also 3.
Because 1*1,3*3 etc are also odd.
So favourable outcomes are 3*3=9.
prob=9/36 ->odd probability,
for prob(even)= 1-prob(odd).
If we are looking at how many odd ways that better and easy to think :
1,3,5 are the odd digits in dice
to get odd both numbers should be odd
so the number of chances for the first number is 3,
and for second also 3.
Because 1*1,3*3 etc are also odd.
So favourable outcomes are 3*3=9.
prob=9/36 ->odd probability,
for prob(even)= 1-prob(odd).
Haridev Purve said:
7 years ago
Odd * Even = Even.
Total=(3+3+3)=9;
Even*odd=even
Total=(6+6+6)=18;
So here total odd number is (1,3,5);
And even (2,4,6);
Now E=27;
And Total=6*6=36
Prob=27/37=3/4 ans.
Total=(3+3+3)=9;
Even*odd=even
Total=(6+6+6)=18;
So here total odd number is (1,3,5);
And even (2,4,6);
Now E=27;
And Total=6*6=36
Prob=27/37=3/4 ans.
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