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 3 of 6.
Prajapati divya h said:
1 decade ago
How do it?
In throws of a dice, n(s) = (6*6) = 36.
In throws of a dice, n(s) = (6*6) = 36.
Sree Tulasi said:
1 decade ago
In that question they have asked for 2 dices, it means each dice is having 6 sides. So for 2 dices= 6*6=36. n(S)=36
Then we should find the possibility for sum of 9 so we can take (4,5),(3,6),(6,3),(5,4). These are 4 possibilities for sum of 9. n(E)=4.
After that Probability of E = n(E)/n(S) = 4/36 = 1/9.
Then we should find the possibility for sum of 9 so we can take (4,5),(3,6),(6,3),(5,4). These are 4 possibilities for sum of 9. n(E)=4.
After that Probability of E = n(E)/n(S) = 4/36 = 1/9.
Rama said:
1 decade ago
How we got n(e) i.e, 4. Please friends somebody help me to find out the answer.
Lettisha L.S said:
1 decade ago
@Rama.
So n(e)=4. It's because the possible combinations where you would get sum 9 when you throw the dice twice are 4.
Like in a dice there are 1-6 no.s right? So what no.s do you put together to get sum 9? (3,6), (4,5), (6,3) ,(5,4) .
But these combos aren't possible: (2,7), (1,8) because well they're not on a dice.
Hope THAT cleared it up for you! ^_^.
So n(e)=4. It's because the possible combinations where you would get sum 9 when you throw the dice twice are 4.
Like in a dice there are 1-6 no.s right? So what no.s do you put together to get sum 9? (3,6), (4,5), (6,3) ,(5,4) .
But these combos aren't possible: (2,7), (1,8) because well they're not on a dice.
Hope THAT cleared it up for you! ^_^.
Nani said:
1 decade ago
How can we get outcomes as 36? Please explain.
Ajax said:
1 decade ago
Short method of find sum of two dice.
Sum - Time.
2 - 1.
3 - 2.
4 - 3.
5 - 4.
6 - 5.
7 - 6.
8 - 5.
9 - 4.
10 - 3.
11 - 2.
12 - 1.
Sum - Time.
2 - 1.
3 - 2.
4 - 3.
5 - 4.
6 - 5.
7 - 6.
8 - 5.
9 - 4.
10 - 3.
11 - 2.
12 - 1.
Reena said:
1 decade ago
Two of a dice n(S) = 6*6 because 1 dice = 1, 2, 3, 4, 5, 6.
1 dice = 1, 2, 3, 4, 5, 6.
= 1(6)*1(6) = 36.
Let E = Event of getting a sum = {(3, 6), (4, 5), (5, 4), (6, 3)}.
= (3+6) = 9.
= (4+5) = 9.
= (5+4) = 9.
= (6+3) = 9.
As a question,
P(E) = n(E)/n(S) = 4/36 = 1/9.
1 dice = 1, 2, 3, 4, 5, 6.
= 1(6)*1(6) = 36.
Let E = Event of getting a sum = {(3, 6), (4, 5), (5, 4), (6, 3)}.
= (3+6) = 9.
= (4+5) = 9.
= (5+4) = 9.
= (6+3) = 9.
As a question,
P(E) = n(E)/n(S) = 4/36 = 1/9.
John said:
1 decade ago
Do you need to do all thus working out?
Laxmipriya said:
10 years ago
Why to do multiplication like 6*6=36? Why we can't do addition?
Like 6+6 = 12. I can't understand when to do addition and when to do multiplication.
Like 6+6 = 12. I can't understand when to do addition and when to do multiplication.
Amol Pawar said:
10 years ago
Dice has 6 faces, i.e (1, 2, 3, 4, 5, 6) in this addition of 9 occurs when (5+4 or 4+5) and (6+3 or 3+6) that means 4 ways.
In this way n(e) = 4 and n(s) = 6*6 =36.
P (e) = n(e)/n(s) = 4/36 = 1/9.
In this way n(e) = 4 and n(s) = 6*6 =36.
P (e) = n(e)/n(s) = 4/36 = 1/9.
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