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.
Sameer said:
5 years ago
Product = multiplication of both dice number is even.
T.K. Dixit said:
6 years ago
The answer should be 1/2.
(4)
ATUL PRAKASH said:
6 years ago
The simple method is : we have 3 odd number(1, 3 ,5) and 3 even number(2, 4, 6);
We know that (even* even=even): = (3 even number * 3 even number=9));
(odd* even=even):= (3 odd number * 3 even number=9);
(even*odd=even):=(3 even number * 3 odd number=9);
Total even sum=9+9+9 = 27,
Total possiblity of two dice are = 6*6 = 36 (because each dice have 6 faces),
Therefore probablity=27/36 = 3/4.
Note: Here I have taken even*odd and odd*even because either of dice may contain even or odd;i.e: 1st dice contain 1 and second will contain 2 or 1st will contain 2 and 2nd will contain 1.
We know that (even* even=even): = (3 even number * 3 even number=9));
(odd* even=even):= (3 odd number * 3 even number=9);
(even*odd=even):=(3 even number * 3 odd number=9);
Total even sum=9+9+9 = 27,
Total possiblity of two dice are = 6*6 = 36 (because each dice have 6 faces),
Therefore probablity=27/36 = 3/4.
Note: Here I have taken even*odd and odd*even because either of dice may contain even or odd;i.e: 1st dice contain 1 and second will contain 2 or 1st will contain 2 and 2nd will contain 1.
(18)
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.
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.
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).
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.
Anusuya said:
8 years ago
I don't understand why take product why only take sum?
Please, guys explain.
Please, guys explain.
Zaheer said:
8 years ago
Is there any shortcut method to solve such types of question.
Naveena said:
8 years ago
I can't understand it please explain this problem.
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