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?
1
6
1
8
1
9
1
12
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 2 of 6.

Pooja gm said:   8 years ago
Why we take this case only?
(3+6)
(4+5)
(5+4)
(6+3).

This also possible;
(7+2),
(1+8),
(2+7),
(8+1),

Please explain the reason.
(1)

Joshua said:   7 years ago
Because in a throw of two dices there is only 6x6, is the last the number won't Or should not exceed 6.
(1)

RRK said:   7 years ago
Is throwing a dice two times and throwing two dice together, same?

Please, anyone, clear me.
(1)

Surya said:   7 years ago
One dice have six faces so here we have two dices, so the formula is n2 (n Square). So 6*6 =36.
(1)

John said:   1 decade ago
Do you need to do all thus working out?

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.

Laxmipriya said:   1 decade 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.

Amol Pawar said:   1 decade 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.

Deepthi said:   1 decade ago
By throwing 2 dies, to get a sum of 9 there are also possibilities of getting (1,8) (2,7) (3,6) (4,5) (5,4) (6,3) (7,2) (8,1) we have 8 possibilities please explain this.

Chethan said:   10 years ago
Why they took n(E)/n(S)? In some other problem, they took n(S)/n(E).


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