Statistics- Rolling a loaded die

  • #1

Homework Statement



A certain die is weighted such that probabilities of showing a 1, 2, 3, 4, 5, and 6 are
(6/34),
(8/34),
(5/34),
(3/34),
(8/34),
and
(4/34)

A) If two such dice are thrown, and you are told that the sum of the two is 10 or larger. What is the probability that the result was a pair of 5's?

B) How many times would you have to throw this die to have the probability of throwing a 2 exceed 40 percent?


Homework Equations



The Attempt at a Solution



A) For this the possible outcomes for a sum of 10 or greater are:

4-6
5-5
5-6
6-4
6-5
6-6

Getting rid of duplicates, since order doesn't matter there is a 1/4 chance it's double fives. I multiplied this by (8/34)(8/34) and got a probability of .01384

B) For this I used the equation

Probability = Successful outcomes/Total number of outcomes

so .4 = (8/34)/n

solving for n I get .588, which doesn't make sense.


I'm just learning statistics, so if anything I tried to do offends you mathematically, I'm very sorry!
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement



A certain die is weighted such that probabilities of showing a 1, 2, 3, 4, 5, and 6 are
(6/34),
(8/34),
(5/34),
(3/34),
(8/34),
and
(4/34)

A) If two such dice are thrown, and you are told that the sum of the two is 10 or larger. What is the probability that the result was a pair of 5's?

B) How many times would you have to throw this die to have the probability of throwing a 2 exceed 40 percent?


Homework Equations



The Attempt at a Solution



A) For this the possible outcomes for a sum of 10 or greater are:

4-6
5-5
5-6
6-4
6-5
6-6

Getting rid of duplicates, since order doesn't matter there is a 1/4 chance it's double fives. I multiplied this by (8/34)(8/34) and got a probability of .01384

B) For this I used the equation

Probability = Successful outcomes/Total number of outcomes

so .4 = (8/34)/n

solving for n I get .588, which doesn't make sense.


I'm just learning statistics, so if anything I tried to do offends you mathematically, I'm very sorry!

Don't apologize; we are not offended by errors. Besides, this is a probability question, not a statistics one.

Anyway, in (a): eliminating duplicates is an error. For example, 6-4 and 4-6 both contribute equally to getting '10'. For (a) I get the answer 8/21 ≈ 0.38095.

For (b), you want to find the smallest n so that the probability of getting at least one '2' in trials 1,2,...,n is = 0.40. I'll just give a hint to get you going. Look at the experiments as having only two outcomes: S (success = get a '2') or F (failure = not a '2'). Let p = 8/34 = 4/17 be the success probability per trial and q = 1-p = 13/17 be the failure probability per trial.

What is the probability that your first S occurs in trial n? For that to happen, the first (n-1) trials must all give F. What is the probability for that to happen? The probability that the first S occurs on or before trial n is a sum of such probabilities.
 
Last edited:
  • #3
Would you mind explaining how you got part A? I can't figure out how you did it.
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Would you mind explaining how you got part A? I can't figure out how you did it.

If S is the sum we have
[tex] P\{(5,5)|S \geq 10\} = \frac{P\{(5,5)\; \& \;S \geq 10 \}}{P\{S \geq 10\}} \\
= \frac{P\{(5,5)\}}{ P\{S \geq 10 \}}[/tex]
because {(5,5) & S ≥ 10} = {(5,5)} (since the event {(5,5)} is a subevent of {S ≥ 10}).
So, you need to compute P{S ≥ 10}.
 

Related Threads on Statistics- Rolling a loaded die

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
2K
Replies
7
Views
8K
  • Last Post
Replies
22
Views
1K
Replies
11
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
53
Views
704
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
4
Views
2K
Top