Aptitude - Probability - Discussion

As we know one dice has 6 sides(4 edges + 1frontside + 1backside = 6)
Possibilities of getting sum 9 are (6,3)(5,4)(4,5)(3,6) i.e 4 possibilities.
Because;
6+3 = 9
5+4 = 9
4+5 = 9
3+6 = 9.

Remember! The numbers you add should not exceed the 6 digit.

Solution for your question is @Dhaval.
4/6 * 6 * 6 = 4/216 or 1/54.
Hope,This Helps!

I have a little bit confusion. Question is "What is the probability of getting a sum 9 from two throws of a dice?". Here we talk about two throws means that a dice is thrown 2 times.

Further it means that a single dice is throw 2 times. So, we should take sample space of 6 because of a single dice.

In case of 36 the question should be "What is the probability of getting a sum 9 when two dice are thrown?". Or perhaps I am in mistake in understanding the question. Please clear this.

One dice means 6 faces.
Two dice means 6 * 6 = 36 faces.
Either,
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
Total = 36 outcomes.
So question ask that,
Sum =9.
So,
3 + 6 = 9 (3,6)
6 + 3 = 9 (6,3)
4 + 5 = 9 (4,5)
5 + 4 = 9 (5,4)
4 out comes.
p = num of outcomes/total outcomes.
p = 4/36 = 1/9.

Hope you understand this.

@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! ^_^.

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

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.

Look in a dice there are number 1, 2, 3, 4, 5, 6.

To form as a 9 we need 6 + 5 + 4 + 3 number. Such as 5 + 4 = 9, 6 + 3 = 9 but 1 + 8 or 2 + 7 is not possible because dice do not contain the number 7, 8.

So sum up the number 6 +5 + 4 + 3 = 18 (one dice). 18 * 2 = 36 (two dice).

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.

@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.

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.