Question 1.4.1
A committee of 4 people is to be chosen from a group of 10 people (6 men and 4 women).
How many different committees of 4 people can be formed?
How many of these committees consist of exactly 2 men and 2 women?
Since the order in which committee members are chosen doesn’t matter, we use combinations. We are choosing 4 people out of 10: \[
\begin{aligned}
\binom{10}{4} &= \frac{10!}{4! \, 6!} \\
&= \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \\
&= \frac{5040}{24} \\
&= 210
\end{aligned}
\] Therefore, there are \(\boldsymbol{210}\) different possible committees.
We need to choose 2 men out of 6, and separately choose 2 women out of 4. By the multiplication principle, the number of committees with exactly 2 men and 2 women is \[
\binom{6}{2} \times \binom{4}{2} = 15 \times 9 = 90
\]
Therefore, there are \(\boldsymbol{90}\) committees consisting of exactly 2 men and 2 women.
Question 1.4.2
Suppose we roll eight fair six-sided dice.
What is the probability that all eight dice show a 6?
What is the probability that all eight dice show the same number?
By independence, the probability that all eight dice show a 6 is \[
\mathbb{P}(\text{all eight show six}) = \left(\frac{1}{6}\right)^8 = \frac{1}{1{,}679{,}616}
\] Therefore, the probability that all eight dice show a 6 is \(\boldsymbol{\dfrac{1}{1{,}679{,}616}}\).
By additivity, summing over the six possible common values: \[
\begin{aligned}
\mathbb{P}(\text{all eight show same}) &= \sum_{i=1}^{6} \mathbb{P}(\text{all eight show } i) \\
&= \sum_{i=1}^{6} \left(\frac{1}{6}\right)^8 \\
&= 6 \left(\frac{1}{6}\right)^8 \\
&= \left(\frac{1}{6}\right)^7 \\
&= \frac{1}{279{,}936}
\end{aligned}
\] Therefore, the probability that all eight dice show the same number is \(\boldsymbol{\dfrac{1}{279{,}936}}\).
Question 1.4.3
Suppose we repeatedly roll two fair six-sided dice, considering the sum of the two values showing each time. What is the probability that the first time the sum is exactly 7 is on the third roll?
The probability of getting a sum of 7 on any one roll is \[
\mathbb{P}(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}
\] For the first sum of 7 to occur on the third roll, we need the first two rolls to not give a sum of 7, and the third roll to give a sum of 7. Since the rolls are independent, \[
\begin{aligned}
\mathbb{P}(\text{first 7 on third roll}) &= \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right) \\
&= \frac{25}{216}
\end{aligned}
\] Therefore, the probability that the first time the sum is exactly 7 is on the third roll is \(\boldsymbol{\dfrac{25}{216}}\).