Probability - Biased Die

[Skelly4444]

A six-sided biased die is thrown and it is twice as likely to land on a number 5 as it is any other number.
All the other numbers are equally likely to occur. The die is thrown repeatedly.

Find the probability that the first 5 will occur on the sixth throw.

I have tried absolutely every permutation here and cannot obtain the answer of 0.0531 in the back of the book.

Completely baffled as to how they've arrived at this answer?

Any guidance would be appreciated.

 

[blamocur]

What are the chances that the first five throws all produce non-5 sides?

 

[Dr.Peterson]

A six-sided biased die is thrown and it is twice as likely to land on a number 5 as it is any other number.
All the other numbers are equally likely to occur. The die is thrown repeatedly.

Find the probability that the first 5 will occur on the sixth throw.

I have tried absolutely every permutation here and cannot obtain the answer of 0.0531 in the back of the book.

Completely baffled as to how they've arrived at this answer?

Any guidance would be appreciated.

First you need to find the probability of throwing a 5 (on a single throw). What is it?

 

[BigBeachBanana]

A six-sided biased die is thrown and it is twice as likely to land on a number 5 as it is any other number.
All the other numbers are equally likely to occur. The die is thrown repeatedly.

Find the probability that the first 5 will occur on the sixth throw.

I have tried absolutely every permutation here and cannot obtain the answer of 0.0531 in the back of the book.

Completely baffled as to how they've arrived at this answer?

Any guidance would be appreciated.

Have you learned about geometric distribution?

 

[Skelly4444]

I modelled it as if the die had the numbers 1 to 4 on it and then two number 5s on the other 2 faces.
This would then give the likelihood of obtaining a 5 as twice that of any other number.

My calculation was then (1/3)^5 x (2/3) which does not agree with the answer in the book. Where am I going wrong?

 

[BigBeachBanana]

I modelled it as if the die had the numbers 1 to 4 on it and then two number 5s on the other 2 faces.
This would then give the likelihood of obtaining a 5 as twice that of any other number.

My calculation was then (1/3)^5 x (2/3) which does not agree with the answer in the book. Where am I going wrong?

This is how I would go about finding the probability of each face value.

The probability of getting 1, 2 ,3, 4, 6 are equally likely so we have
[imath]\Pr(X=1)=\Pr(X=2)=\Pr(X=3)=\Pr(X=4)=\Pr(X=6)=p[/imath]

Since the face-value 5 is twice likely, you'd get [imath]\Pr(X=5)=2p[/imath]

The total probability is [imath]p+p+p+p+p+2p=7p[/imath]

What is the probability of getting a 5 on a single throw, i.e. [imath]\Pr(X=5)[/imath]?

 

[Skelly4444]

I think I've worked out where I'm going wrong.
The probability distribution gives 7x = 1 , giving probabilities of 1/7 and 2/7.

Then I keyed in (5/7)^5 x (2/7) and this is the same answer as the back of the book.

Thank you all for your hel

 

[Steven G]

Skelly,
Your biggest problem was assuming that since the chances of getting a 5 was twice as likely as getting another number that this means that you replace the 6 (or 1, 2, 3, 4) with a 5. This would mean that the chance of rolling a 6 is 0. Why should that be? You are correct that in this case it is like having an extra 5, but not at the expense of removing another number. You can and probably think of this die as having 7 faces.