Probabilities, need help test is tommorow.

[thelazyman]

Hi, I have two questions about permutations and probabilities.

The question for both is: Determine the probability of each of the following situations:

a) at least one 3 turns up when 3 dice are rolled
b) a committee of six people randomly chosen from seven males and eight females is either all male or all female.

This is what i tried doing so far.

For a) i tried to determine its 1/6 X 5/6 X5/6, but that was a flop and didnt go so far.


For b) i tried doing:

7!/15!/(15-6)! + 8!/15!/(15-6)! but the answer for both were wrong..


Please help

 

[pka]

The probability that NO THREES is \(\displaystyle \L
\left( {\frac{5}{6}} \right)^3\) .
So the probability of at least one three is \(\displaystyle \L
1 - \left( {\frac{5}{6}} \right)^3\) .

All of one sex: \(\displaystyle \L
\frac{{\left( \begin{array}{c}
7 \\
6 \\
\end{array} \right) + \left( \begin{array}{c}
8 \\
6 \\
\end{array} \right)}}{{\left( \begin{array}{c}
15 \\
6 \\
\end{array} \right)}}\)