# Thread: dice probability question

1. ## dice probability question

A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is
four times the probability of getting 2 even numbers. Find the probability that a single throw of the
die results in an even number.

Thanks

2. Originally Posted by ipokeyou
A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is
four times the probability of getting 2 even numbers. Find the probability that a single throw of the
die results in an even number.

Thanks
Hi ipokeyou,

This is a binomial situation.
We either get an even number or an odd number on any throw.
We don't need to examine the specific 6 numbers themselves.

There are $\binom{8}{3}$ ways to get 3 even outcomes in 8 throws (doesn't matter which even number) and $\binom{8}{2}$ ways to get 2 even outcomes.

Therefore

$\binom{8}{3}p^3(1-p)^5=4\binom{8}{2}p^2(1-p)^6$

since 3 even throws must go with 5 odd throws etc
and p=probability of an even result, 1-p=probability of an odd result.

Hence,

$\frac{8!}{5!3!}p^3(1-p)^5=4\frac{8!}{6!2!}p^2(1-p)^6$

Diving both sides by $p^2(1-p)^5$

$\frac{8(7)6}{6}p=\frac{4(8)7}{2}(1-p)$

$56p=112(1-p)$

$56p=112-112p$

$168p=112$

$p=\frac{112}{168}=\frac{28}{42}=\frac{4}{6}=\frac{ 2}{3}$

3. thanks for your help and quick response!!

4. Hello, ipokeyou!

A slight variation . . .

A die is loaded in such a way that in 8 throws of the die, the probability
of getting 3 even numbers is four times the probability of getting 2 even numbers.
Find the probability that a single throw of the die results in an even number.

Let: . $\begin{array}{ccc}p &=& P(\text{even}) \\ q &=& P(\text{odd}) \end{array}\quad\text{ where }p+q\,=\,1$

We have: . $\begin{array}{ccccc}P(\text{3 even)} &=& {8\choose3}p^3q^5 &=& 56p^3q^5\\ \\[-3mm]P(\text{2 even}) &=& {8\choose2}p^2q^6 &=& 28p^2q^6\end{array}$

. . Then: . $56p^3q^5 \:=\:4\cdot28p^2q^6 \quad\Rightarrow\quad p \:=\:2q$

Since $p+q\:=\:1$, we have: . $p = \tfrac{2}{3},\;q =\tfrac{1}{3}$

Therefore: . $P(\text{even}) \:=\:\frac{2}{3}$