Probability (1 Viewer)

[swagswagyoloswag]

An unbiased die is thrown six times. Find the probability that the six scores obtained will be such that a 6 occurs only on the last throw and exactly three of the first five throws result in odd numbers.
How would I start this?

 

[InteGrand]

An unbiased die is thrown six times. Find the probability that the six scores obtained will be such that a 6 occurs only on the last throw and exactly three of the first five throws result in odd numbers.
How would I start this?

Pr(6 on last throw) = 1/6.

We require now exactly three of the first five to have odd numbers (so exactly two have an even number that is not 6, i.e. 2 or 4. Also, Pr(2 or 4 on a given roll) = 2/6 = 1/3).

There are ways to pick to the throws that come up with an even number. Having picked these, we require them both to have 2 or 4 (so probability of this is ), and we require the other three to also be odd (probability of this is , since Pr(odd) = 1/2). So the probability of exactly three of the first five throws being odd scores AND the other two being 2 or 4 is .

So the overall answer is .

 

[swagswagyoloswag]

What was wrong with how I did it?
[1] [2] [3] [4] [5] [6]
For [1],[2],[3] i want 1,3,5 so
Then [4],[5] , it can be any number but 6 so
Then [6] I only want 6, so p =
Therefore it is
.
Another method I did was for the first three, there are 3x3x3 = 27 possible choices. Then for 4th,5th, there are 5x5 = 25 choices and for 6th, there is 1 choice. So there is 675 possible. Sample space = 6^6
This gave me same answer as method 1

 

[InteGrand]

What was wrong with how I did it?
[1] [2] [3] [4] [5] [6]
For [1],[2],[3] i want 1,3,5 so
Then [4],[5] , it can be any number but 6 so
Then [6] I only want 6, so p =
Therefore it is
.
Another method I did was for the first three, there are 3x3x3 = 27 possible choices. Then for 4th,5th, there are 5x5 = 25 choices and for 6th, there is 1 choice. So there is 675 possible. Sample space = 6^6
This gave me same answer as method 1

Both the above methods seem to assume that the odd-numbered scores need to occur in the first three rolls. But the question says they can occur in any three of the first five rolls.

 

[swagswagyoloswag]

Would I have to multiply by 3!?

 

[InteGrand]

What was wrong with how I did it?
[1] [2] [3] [4] [5] [6]
For [1],[2],[3] i want 1,3,5 so

Would I have to multiply by 3!?

Why are you doing ? That part should just be . And then you need to multiply by to account for the different allowable orderings of where your odd-numbered scores can go.

 

[swagswagyoloswag]

Why are you doing ? That part should just be . And then you need to multiply by to account for the different allowable orderings of where your odd-numbered scores can go.

THat means 5C3 right? I understand that you have to have that but I can't think of a logical reason. can you give an example using that? Im not good with perms/combs and probability

 

[InteGrand]

THat means 5C3 right? I understand that you have to have that but I can't think of a logical reason. can you give an example using that? Im not good with perms/combs and probability

Yeah, 5C3 is the number of ways we can order the three odd-numbered results from the first five rolls. We can have them in any three of the first five rolls, e.g. in roll #'s {1,3,5}, {2,3,4}, {1,2,4} etc. In order to account for each of these possibilities (which are equally likely), we have to multiply the (which is the probability of a GIVEN set of three rolls having all numbers odd) by the number of such sets of three rolls (and this number is , since that's the number of ways of choosing three things from a group of five; in this case, we chose a set of three rolls from a group of five rolls).

 

[swagswagyoloswag]

Yeah, 5C3 is the number of ways we can order the three odd-numbered results from the first five rolls. We can have them in any three of the first five rolls, e.g. in roll #'s {1,3,5}, {2,3,4}, {1,2,4} etc. In order to account for each of these possibilities (which are equally likely), we have to multiply the (which is the probability of a GIVEN set of three rolls having all numbers odd) by the number of such sets of three rolls (and this number is , since that's the number of ways of choosing three things from a group of five; in this case, we chose a set of three rolls from a group of five rolls).

I still dont really get why its 5C3 instead of just multiplying by 3

 

[InteGrand]

I still dont really get why its 5C3 instead of just multiplying by 3

This is because there's 5C3 different ways of ordering the three odd-number in the first five rolls. It's just like how the number of ways of picking three doors to open from a set of 5 closed doors is 5C3. In this question, instead of opening doors, we are choosing which rolls will land an odd.

 

[Ekman]

I still dont really get why its 5C3 instead of just multiplying by 3

Here's a different way of looking at it:

o,e,o,o,e Where e=even and o=odd
And so the number of arrangements of throws is: 5!/2!*3! (since all odds and evens are identical) = 5C3 or 5C2

 

[swagswagyoloswag]

This is because there's 5C3 different ways of ordering the three odd-number in the first five rolls. It's just like how tghe number of ways of picking three doors to open from a set of 5 closed doors is 5C3. In this question, instead of opening doors, we are choosing which rolls will land an odd.

ohh thanks that example helped