Aptitude - Probability - Discussion
Discussion Forum : Probability - General Questions (Q.No. 4)
4.
What is the probability of getting a sum 9 from two throws of a dice?
Answer: Option
Explanation:
In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
 P(E) = |
n(E) | = | 4 | = | 1 | . |
| n(S) | 36 | 9 |
Discussion:
52 comments Page 4 of 6.
Krish said:
1 decade ago
How it could be n(s)= (6 x 6)
Srinivas said:
1 decade ago
n(s)=36
n(a)=[36][45][63][54]
p(a)=4/36=1/9
n(a)=[36][45][63][54]
p(a)=4/36=1/9
Sreejith said:
1 decade ago
@Azam
In both throughs, it can be any of 1,2,3...,6(6C1 ways of selection). So total number of outcomes = 6C1*6C1 = 36
More clearly, the elements of sample spaces are
{(1,1),(1,2),(1,3),....,(2,1),(2,2),.....(6,1),(6,2),....(6,6)}
Hope you got it.
In both throughs, it can be any of 1,2,3...,6(6C1 ways of selection). So total number of outcomes = 6C1*6C1 = 36
More clearly, the elements of sample spaces are
{(1,1),(1,2),(1,3),....,(2,1),(2,2),.....(6,1),(6,2),....(6,6)}
Hope you got it.
Harikannan said:
1 decade ago
How to find event that is (3, 4) (4, 5) (5, 4) (6, 3) please explain.
Syam said:
1 decade ago
{ (3, 6) , (4, 5) , (5, 4) , (6, 3) } is a case only.
We can get (6, 3) , (5, 4) , (3, 6) , (4, 5) also.
So answer is 8/38 = 2/9.
Am I correct ?
We can get (6, 3) , (5, 4) , (3, 6) , (4, 5) also.
So answer is 8/38 = 2/9.
Am I correct ?
Ankita said:
1 decade ago
How do we know that we have to pick up 6 * 6?
Anurag said:
1 decade ago
Ankita:.
Because 6c1 * 6c1 is 36.
Because 6c1 * 6c1 is 36.
Kanchi said:
1 decade ago
s is a sample space i.e all possible outcomes
if two coins are tossed then s={HH,TT,HT,TH}
same in this qu. two dice -use then no of possible outcome
s={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)},
{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}
..........
{(6,1),(6,2),(6,3),(6,4),(6,5),(6,5)}
n(s)=6+6+6+6+6+6=36
or
6*6=36
if two coins are tossed then s={HH,TT,HT,TH}
same in this qu. two dice -use then no of possible outcome
s={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)},
{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}
..........
{(6,1),(6,2),(6,3),(6,4),(6,5),(6,5)}
n(s)=6+6+6+6+6+6=36
or
6*6=36
Frederick Mattson said:
1 decade ago
1,1 2,1 3,1 4,1 5,1 6,1
1,2 2,2 3,2 4,2 5,2 6,2
1,3 2,3 3,3 4,3 5,3 6,3
1,4 2,4 3,4 4,4 5,4 6,4
1,5 2,5 3,5 4,5 5,5 6,5
1,6 2,6 3,6 4,6 5,6 6,6
P(sum 9) = {(6,3),(5,4),(4,5),(3,6)}.
= 4/36.
= 1/9.
1,2 2,2 3,2 4,2 5,2 6,2
1,3 2,3 3,3 4,3 5,3 6,3
1,4 2,4 3,4 4,4 5,4 6,4
1,5 2,5 3,5 4,5 5,5 6,5
1,6 2,6 3,6 4,6 5,6 6,6
P(sum 9) = {(6,3),(5,4),(4,5),(3,6)}.
= 4/36.
= 1/9.
Malasrvizhi said:
1 decade ago
How did get the sum element?
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