What's the probability of getting a total of six points or less when throwing 3 dices

[deah]

what is the probability of getting a total of six points or less when
throwing 3 dice?

 

[A.Anand Srinivasan]

Re: probabilities!!!!

it is 20/216.....

 

[deah]

Re: probabilities!!!!

what will be my process in taking that answer?

 

[A.Anand Srinivasan]

Re: probabilities!!!!

find the total number of possible outcomes with the dices which is 216.... now find the number of combinations that give a sum of 3,4,5,6.... it will be 1+3+6+10=20 and hence the probability is 20/216...

 

[Dmitrij]

Re: probabilities!!!!

You must also take into consideration? that the same combination may occur with different numbers on every dice. For example;

1 1 4 (6 points)
4 1 1 (6 points as well)
1 4 1 (6 points as well)

So the amount of combinations should be multipled by 3:

N = 10 + 6 + 3 + 1 = 20

P = 20/216

With respect,

Dmitrij

 

[deah]

Re: probabilities!!!!

how about..
A number is chosen at random between 0 and 1.
what is the probability that exactly 5 of its first ten decimal places consist of digits less than 5?
and
consider a game in which 6 true dice are rolled. find the probability of obtaining
a) exactly one ace
b)at least one ace
c) exactly two aces

thank you.....

 

[A.Anand Srinivasan]

Re: probabilities!!!!

for the first case the answer is ((5^5)*(10^5))/((10^10)-1) is the answer for the first case.....

for the second case i dont know what you mean by ace in a dice.... please specify it...

 

[deah]

Re: probabilities!!!!

oh, i'm sorry..
ace means one..

 

[A.Anand Srinivasan]

Re: probabilities!!!!

for the dice problem
a)1/6
b)1-((5^6)/(6^6))
c)15/36

 

[Dmitrij]

Re: probabilities!!!!

Concerning the first proble, my result is the following:

(5^5)*(10^5) / (10^10)

My formula differs from the A.Anand Srinivasan's one, concerning denominator. I explain it by the fact, that there are 10 positions in the decimal part of the number. Each position in general may be occupied by any of 10 digits (0,1,2,....,9), therefore the total number of all possible states is 10*10*...*10 = 10^10.

Please, if you are exactly aware of the correct answer, let me know. I'm particularly curious and anxious about the proclaimed solution.

With respect,

Dmitrij

Added after 20 minutes:

The second problem must be specified in some aspects. However, I won't irritate you by permant requests of formulating the condition, but I'll offer my solution.

Take into consideration, that I consider the events of rolling every dice - independent and the possible values of the dice's score: 1,2,3,4,5,6 (INSTEAD OF 0,1,2,3,4,5).

1) p = 1/(6^6) (6^6 - THE TOTAL NUMBER OF DIFFERENT RESULTS DURING THE ROLLING PROCESS)

2) p = 1 (because the minimum score on all dices is 1, it can't be less, according to my preliminary remark. so you'll always get at least 1)

3) 15/6^6

With respect,

Dmitrij

 

[A.Anand Srinivasan]

Re: probabilities!!!!

Concerning the first proble, my result is the following:

(5^5)*(10^5) / (10^10)

My formula differs from the A.Anand Srinivasan's one, concerning denominator. I explain it by the fact, that there are 10 positions in the decimal part of the number. Each position in general may be occupied by any of 10 digits (0,1,2,....,9), therefore the total number of all possible states is 10*10*...*10 = 10^10.

Please, if you are exactly aware of the correct answer, let me know. I'm particularly curious and anxious about the proclaimed solution.

With respect,

Dmitrij

Added after 20 minutes:

The second problem must be specified in some aspects. However, I won't irritate you by permant requests of formulating the condition, but I'll offer my solution.

Take into consideration, that I consider the events of rolling every dice - independent and the possible values of the dice's score: 1,2,3,4,5,6 (INSTEAD OF 0,1,2,3,4,5).

1) p = 1/(6^6) (6^6 - THE TOTAL NUMBER OF DIFFERENT RESULTS DURING THE ROLLING PROCESS)

2) p = 1 (because the minimum score on all dices is 1, it can't be less, according to my preliminary remark. so you'll always get at least 1)

3) 15/6^6

With respect,

Dmitrij

for the first problem the case of .0000000000 doesnt fall between 0 and 1 and hence we'll have only ((10^10)-1) combinations.....

for the second problem
b) it means that atleast one of the face of the 6 dices show a one and how many 1's do you think the combination 6,6,6,6,6,6 has..... that probability cannot be 1
c) you haven't considered the fact that the other 4 faces of the dice can come in combinations of their own....

 

[Dmitrij]

Re: probabilities!!!!

Concerning the 1st problem, you're right, because I've forgotten about the special case .0000000000. Thus, the answer is:

p = (5^5)*(10^5) / (10^10 - 1)

In the 2nd problem:

c) I think I really haven't taken into consideration the possible combinations of the rest dices, except 2 with the score 1. We get *11***, for example. In general there are 15 different combinations with two 1. The total amount of the possible results is 6^6. So we get

p = 15*(5^4)/(6^6)

If you disagree with my suggestion, please point your own solution and arguments to confirm your opinion. We'll compare.

b) p = 1 - (5^6)/(6^6)

a) p = (5/6)^5

With respect,

Dmitrij