# Thread: probability of throwing a die

1. ## probability of throwing a die

A die is weighted so that the probability of its landing face i uppermost is i/21. This die is thrown and if i shows uppermost, i fair dice are thrown
(i = 1,2,3,4,5,6). A total score is obtained by summing the numbers showing uppermost on these fair dice.

1) Find the probability that the total score is 4
2) Find the probability that 2 fair dice were thrown, given that the total score is 4.

At the first attempt of this question, I got there are 4 ways which you can a total score of 4 when you roll i dice.
When throwing: 1 die gives only one choice, that is number 4
2 dice give (1,3),(2,2),(3,1) which the sum is 4
3 dice give (1,1,2),(1,2,1),(2,1,1) which the sum is 4
4 dice give (1,1,1,1)
Cuz I am not allowed to use tree diagram, so I have to use the formulae of probability

I've just done up to here then I got stuck, I think somehow we have to use the law of total probability.

Can some body help me to do part 1 and give hints on part 2 please?

Thank you very very much for your time.

2. Originally Posted by knguyen2005
A die is weighted so that the probability of its landing face i uppermost is i/21. This die is thrown and if i shows uppermost, i fair dice are thrown (i = 1,2,3,4,5,6). A total score is obtained by summing the numbers showing uppermost on these fair dice.
1) Find the probability that the total score is 4
2) Find the probability that 2 fair dice were thrown, given that the total score is 4.
If I have understood what you have written, here is part (a):
$\displaystyle \begin{array}{rcl} {P(S = 4)} & = & {\sum\limits_{k = 1}^4 {P(S = 4|i = k)P(i = k)} } \\ {} & = & {\left( {\frac{1}{6}} \right)\left( {\frac{1}{{21}}} \right) + \left( {\frac{3}{{6^2 }}} \right)\left( {\frac{2}{{21}}} \right) + \left( {\frac{3}{{6^3 }}} \right)\left( {\frac{3}{{21}}} \right) + \left( {\frac{1}{{6^4 }}} \right)\left( {\frac{4} {{21}}} \right)} \\ \end{array}$.

For part (b) use Bayes rule.

3. Originally Posted by Plato
If I have understood what you have written, here is part (a):
$\displaystyle \begin{array}{rcl} {P(S = 4)} & = & {\sum\limits_{k = 1}^4 {P(S = 4|i = k)P(i = k)} } \\ {} & = & {\left( {\frac{1}{6}} \right)\left( {\frac{1}{{21}}} \right) + \left( {\frac{3}{{6^2 }}} \right)\left( {\frac{2}{{21}}} \right) + \left( {\frac{3}{{6^3 }}} \right)\left( {\frac{3}{{21}}} \right) + \left( {\frac{1}{{6^4 }}} \right)\left( {\frac{4} {{21}}} \right)} \\ \end{array}$.

For part (b) use Bayes rule.
Plato,

I think some of the values in that last equation are incorrect, given that the probability of rolling an x with one die is x/21:

$\displaystyle P(S=4 | i = 1) = P(4) = 4/21$

$\displaystyle P(S=4 | i=2) = P(1,3) + P(2,2) + P(3,1)$ $\displaystyle = (1/21)(3/21) + (2/21)(2/21) + (3/21)(1/21) = 10 / 21^2$

$\displaystyle P(S=4 | i=3) = P(1,1,2) + P(1,2,1) + P(2,1,1)$ $\displaystyle = 3 (1/21)(1/21)(2/21) = 6/21^3$

$\displaystyle P(S=4 | i=4) = P(1,1,1,1) = 1/21^4$

4. Originally Posted by awkward
I think some of the values in that last equation are incorrect, given that the probability of rolling an x with one die is x/21:
$\displaystyle P(S=4 | i = 1) = P(4) = 4/21$
Frankly, I find you confusion mystifying. You really need to rethink what you have posted.
The probability that a one appears on this bias die is $\displaystyle \frac{1}{21}$.
Tossing one unbiased die and getting a 4 is $\displaystyle \frac{1}{6}$.
So what is the probability of tossing one bias die and getting a 1 then tossing a 4 on one unbiased die?

Where do you get $\displaystyle \frac{4}{21}$???

Given the quality of your other postings, I am confused by this.

5. Originally Posted by Plato
Frankly, I find you confusion mystifying. You really need to rethink what you have posted.
The probability that a one appears on this bias die is $\displaystyle \frac{1}{21}$.
Tossing one unbiased die and getting a 4 is $\displaystyle \frac{1}{6}$.
So what is the probability of tossing one bias die and getting a 1 then tossing a 4 on one unbiased die?

Where do you get $\displaystyle \frac{4}{21}$???

Given the quality of your other postings, I am confused by this.
Oops, you are right! I misread the problem.

I thought we were rolling the biased dice all along, but I now see that after the first roll we are rolling fair dice.

Confused by the old switcheroo!