# Another tricky problem

• Oct 8th 2009, 08:18 AM
multivariablecalc
Another tricky problem
1. Three dice are rolled twice. What is the probability that they show the same number if the dice are distinguishable, and if they are not.

2. One Hundred Trout are caught and returned after they are tagged. Later another 100 are caught and found to contain 7 tagged ones. What is the probability of this if the lake contains n trout? What is your guess at to the true value of n?

these both came off very tricky to me. Can someone take the time to explain? thanks in advance!
• Oct 8th 2009, 01:31 PM
Soroban
Hello, multivariablecalc!

No one is responding because these are particularly ugly problems.

Quote:

Three dice are rolled twice.
What is the probability that they show the same number?

It doesn't matter whether the dice are distinguishable or not.
The probabilties are identical.

I will assume that "show the same number" means:
. . the total of the three dice on the second roll equals
. . the total of the three dice on the first roll.

With three dice, the sums range from 3 to 18.
Here are their probabilities:

. . $\displaystyle \begin{array}{cc}\text{Sum} & \text{Prob} \\ \hline \\[-4mm] 3 & \frac{1}{216} \\ \\[-4mm] 4 & \frac{3}{216} \\ \\[-4mm] 5 & \frac{6}{216} \\ \\[-4mm] 6 & \frac{10}{216} \\ \\[-4mm] 7 & \frac{15}{216} \\ \\[-4mm] 8 & \frac{21}{216} \\ \\[-4mm] 9 & \frac{25}{216} \\ \\[-4mm] 10 & \frac{27}{216} \end{array}$ . . . $\displaystyle \begin{array}{cc}\text{Sum} & \text{Prob} \\ \hline \\[-4mm] 11 & \frac{27}{216} \\ \\[-4mm] 12 & \frac{25}{216} \\ \\[-4mm] 13 & \frac{21}{216} \\ \\[-4mm] 14 & \frac{15}{216} \\ \\[-4mm] 15 & \frac{10}{216} \\ \\[-4mm] 16 & \frac{6}{217} \\ \\[-4mm] 17 & \frac{3}{216} \\ \\[-4mm] 18 & \frac{1}{216} \end{array}$

Then: .$\displaystyle P(\text{same number}) \;=\;P(\text{3 and 3})+ P(\text{4 and 4}) + P(\text{5 and 5}) + \hdots$ .$\displaystyle + P(\text{18 and 18})$

. . . . . . . . . . . . . . . . $\displaystyle = \;\left(\frac{1}{216}\right)^2 + \left(\frac{3}{216}\right)^2 + \left(\frac{6}{216}\right)^2 + \hdots + \left(\frac{1}{216}\right)^2$

I'll let you crank it out . . .

• Oct 8th 2009, 06:08 PM
awkward
Quote:

Originally Posted by multivariablecalc
1. Three dice are rolled twice. What is the probability that they show the same number if the dice are distinguishable, and if they are not.

2. One Hundred Trout are caught and returned after they are tagged. Later another 100 are caught and found to contain 7 tagged ones. What is the probability of this if the lake contains n trout? What is your guess at to the true value of n?

these both came off very tricky to me. Can someone take the time to explain? thanks in advance!

I'll try #2.

I'm assuming the second sample is taken without replacement. If the lake contains n fish, of which 100 are tagged and n-100 are not, then the probability that 7 tagged fish will be found in a sample of size 100 is
$\displaystyle \frac{\binom{100}{7} \binom{n-100}{93}}{\binom{n}{100}}$.

As for guessing the true value of n, I am sure this can be done in many ways, but you might ask this question: Suppose we took a sample of 100 fish with replacement. If the mean number of tagged fish in the sample is 7 and the probability that a fish is tagged is 100/n, what must be the value of n?