Probability Question (1 Viewer)

[taco man]

Hi could someone do this question for me please:
A die is thrown 6 times:
-what is the probability that the nth throw is n on exactly 5 occasions?

Thx

 

[word.]

This isn't mathematics level:-
P(nth throw is n on exactly 5 occasions) = ⁶C₅(5/6)(1/6)⁵

 

[insert-username]

-what is the probability that the nth throw is n on exactly 5 occasions?

The probability that it will be 1 on the first throw is 1/6. Same for the 2nd throw, the 3rd throw, and so on. The probability that it will not be the number the last time is 5/6. So, your probability is:

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 5/6

= 5/46656


I_F

 

[klaw]

This isn't mathematics level:-
P(nth throw is n on exactly 5 occasions) = ⁶C₅(5/6)(1/6)⁵

Binomial probability is just a faster way of doing 2U probability

 

[klaw]

2U way:
P(nth throw is n on exactly 5 occasions) =1/6.1/6.1/6.1/6.1/6.5/6+1/6.1/6.1/6.1/6.5/6.1/6+1/6.1/6.1/6.5/6.1/6.1/6+1/6.1/6.5/6.1/6.1/6.1/6+1/6.1/6.5/6.1/6.1/6.1/6+1/6.5/6.1/6.1/6.1/6.1/6+5/6.1/6.1/6.1/6.1/6.5/6 =6(5/6)(1/6)⁵ =5/7776

 

[taco man]

2U way:
P(nth throw is n on exactly 5 occasions) =1/6.1/6.1/6.1/6.1/6.5/6+1/6.1/6.1/6.1/6.5/6.1/6+1/6.1/6.1/6.5/6.1/6.1/6+1/6.1/6.5/6.1/6.1/6.1/6+1/6.1/6.5/6.1/6.1/6.1/6+1/6.5/6.1/6.1/6.1/6.1/6+5/6.1/6.1/6.1/6.1/6.5/6 =6(5/6)(1/6)⁵ =5/7776

can u explain ur answer, i dun relli get how it works. Basically I got the same working out and answer as Insert-Username.

 

[klaw]

can u explain ur answer, i dun relli get how it works. Basically I got the same working out and answer as Insert-Username.

Possible throws:
1st throw: 1 (1/6 chance)
2nd throw: 2 (1/6)
3rd throw: 3 (1/6)
4th throw: 4 (1/6)
5th throw: 5 (1/6)
6th throw: anything but 6 (5/6)

1st throw: 1 (1/6 chance)
2nd throw: 2 (1/6)
3rd throw: 3 (1/6)
4th throw: 4 (1/6)
5th throw: anything but 5 (5/6)
6th throw: 6 (1/6)

1st throw: 1 (1/6 chance)
2nd throw: 2 (1/6)
3rd throw: 3 (1/6)
4th throw: anything but 4 (5/6)
5th throw: 5 (1/6)
6th throw: 6 (1/6)

1st throw: 1 (1/6 chance)
2nd throw: 2 (1/6)
3rd throw: anything but 3 (5/6)
4th throw: 4 (1/6)
5th throw: 5 (1/6)
6th throw: 6 (1/6)

1st throw: 1 (1/6 chance)
2nd throw: anything but 2 (5/6)
3rd throw: 3 (1/6)
4th throw: 4 (1/6)
5th throw: 5 (1/6)
6th throw: 6 (1/6)

1st throw: anything but 1 (5/6 chance)
2nd throw: 2 (1/6)
3rd throw: 3 (1/6)
4th throw: 4 (1/6)
5th throw: 5 (1/6)
6th throw: 6 (1/6)

P(throw n on the nth throw exactly 5 times)=sum of the probability of all those =1/6.1/6.1/6.1/6.1/6.5/6+1/6.1/6.1/6.1/6.5/6.1/6+1/6.1/6.1/6.5/6.1/6.1/6+1/6.1/6.5/6.1/6.1/6.1/6+1/6.1/6.5/6.1/6.1/6.1/6+1/6.5/6.1/6.1/6.1/6.1/6+5/6.1/6.1/6.1/6.1/6.5/6 =6(5/6)(1/6)⁵ =5/7776

 

[taco man]

nice i get it now thanks klaw

 

[insert-username]

Yeah, my answer misses out the other ways that the event can happen - I only worked it out for getting the first five throws.


I_F