Events A and B are such that. P(A) = 1/3, P(B/A)=1/4 and P(A' B') = 1/2. Find P(A B), P(B).

Player A wins a game if:

1. A wins outright: $P(W_A) = \frac{1}{2}$
2. A draws and wins the coin toss: $P(D_A) \times 0.5 = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ So, the total probability that A wins a game (and scores 6 points) is:

Player A loses a game if:

1. A loses outright: $P(L_A) = \frac{1}{6}$
2. A draws and loses the coin toss: $P(D_A) \times 0.5 = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ So, the total probability that A loses a game (and scores 0 points) is:

Note that $P(\text{A wins a game}) + P(\text{A loses a game}) = \frac{2}{3} + \frac{1}{3} = 1$.

The competition is won by the first player to score 6 points or more. Since a win in a game gives 6 points and a loss gives 0 points (after draw resolution), the competition is decided in the first game. If A wins the first game, A scores 6 points and wins the competition. If A loses the first game, B scores 6 points and wins the competition.
Therefore, the probability that A wins the competition is the probability that A wins the first game.

: Geometric Distribution

a) State two conditions necessary for X to have a Geometric distribution.

1. Each attempt is independent of the others.
2. The probability of success ($p$) is constant for each attempt.

b) Assuming that X has the distribution Geo(0.3), find
Here, $p = 0.3$. The probability mass function for a Geometric distribution is $P(X=k) = (1-p)^{k-1} p$.

(i) $P(X=5)$
This is the probability that the first success occurs on the 5th attempt.

(ii) $P(X \ge 6)$
This is the probability that the first success occurs on the 6th attempt or later. This means the first 5 attempts must all be failures.

c) Suggest a reason why one of the conditions you have given in part (a) might not be satisfied in this context.
One condition that might not be satisfied is that the probability of success is constant. Andy might learn from previous attempts, increasing his probability of success, or become fatigued/frustrated, decreasing his probability of success. This would violate the assumption of a constant $p$.

: Choir Selection and Arrangement

The choir consists of: 13 sopranos, 12 altos, 6 tenors, 7 basses.
The group to be chosen consists of: 10 sopranos, 9 altos, 4 tenors, 4 basses.

a) In how many different ways can the group be chosen?
This is a combination problem for each section of the choir.

Ways to choose sopranos: $\binom{13}{10}$
Ways to choose altos: $\binom{12}{9}$
Ways to choose tenors: $\binom{6}{4}$
Ways to choose basses: $\binom{7}{4}$

Total ways = $286 \times 220 \times 15 \times 35 = \boxed{33100000}$

b) In how many ways can the 10 chosen sopranos be arranged in a line if the 6 tallest stand next to each other?

We have 10 sopranos. If 6 tallest stand together, we consider them as one unit. The remaining $10-6=4$ sopranos are individual units.
So, we are arranging 1 block and 4 individual sopranos, which is a total of $1+4=5$ units.

These 5 units can be arranged in $5!$ ways.

The 6 tallest sopranos within their block can be arranged in $6!$ ways.

Total arrangements = (arrangements of units) $\times$ (arrangements within the block)

c) The 4 tenors and 4 basses in the group stand in a single line with all the tenors next to each other and all the basses next to each other. How many possible arrangements are there if three of the tenors refuse to stand next to any of the basses.
Let the 4 tenors be $T_1, T_2, T_3, T_4$ and the 4 basses be $B_1, B_2, B_3, B_4$.
Let the three specific tenors who refuse to stand next to basses be $T_A, T_B, T_C$. The fourth tenor is $T_D$.

Treat the 4 tenors as one block ($T_{block}$) and the 4 basses as one block ($B_{block}$).
These two blocks can be arranged in $2!$ ways ($T_{block} B_{block}$ or $B_{block} T_{block}$).
Within the $T_{block}$, the 4 tenors can be arranged in $4!$ ways.
Within the $B_{block}$, the 4 basses can be arranged in $4!$ ways.
Total arrangements = $2! \times 4! \times 4! = 2 \times 24 \times 24 = 1152$.

The condition is violated if one of the three specific tenors ($T_A, T_B, T_C$) is at an end of the tenor block and is adjacent to a bass block.
Consider the arrangement $T_{block} B_{block}$. The tenor at the right end of $T_{block}$ is adjacent to $B_{block}$.
For the condition to be violated, this tenor must be one of $T_A, T_B, T_C$.
Number of ways to choose this end tenor: 3 (from $T_A, T_B, T_C$).
Number of ways to arrange the remaining 3 tenors in the $T_{block}$: $3!$.
Number of ways to arrange the 4 basses in the $B_{block}$: $4!$.
So, arrangements of type $T_{block} B_{block}$ violating the condition: $3 \times 3! \times 4! = 3 \times 6 \times 24 = 432$.

Consider the arrangement $B_{block} T_{block}$. The tenor at the left end of $T_{block}$ is adjacent to $B_{block}$.
For the condition to be violated, this tenor must be one of $T_A, T_B, T_C$.
Number of ways to choose this end tenor: 3 (from $T_A, T_B, T_C$).
Number of ways to arrange the remaining 3 tenors in the $T_{block}$: $3!$.
Number of ways to arrange the 4 basses in the $B_{block}$: $4!$.
So, arrangements of type $B_{block} T_{block}$ violating the condition: $3 \times 3! \times 4! = 3 \times 6 \times 24 = 432$.

Total arrangements violating the condition = $432 + 432 = 864$.

Number of possible arrangements satisfying the condition = Total arrangements - Violating arrangements

: Sample Mean Probability

Given:
Population mean $\mu = 2$ mm
Population standard deviation $\sigma = 0.05$ mm
Sample size $n = 25$

We need to find the probability that the sample mean ($\bar{X}$) is within 3 standard deviations of the mean. This means:

For $\bar{X} = 1.97$:

For $\bar{X} = 2.03$:

We need to find $P(-3 < Z < 3)$.

From standard normal tables:
$P(Z < 3) \approx 0.99865$
$P(Z < -3) \approx 0.00135$

: Probability of Events and Letters

8a) Events A and B are such that $P(A) = \frac{1}{3}$, $P(B|A) = \frac{1}{4}$ and $P(A' \cap B') = \frac{1}{2}$. Find

(i) $P(A \cup B)$

We know that $P(A' \cap B') = P((A \cup B)')$.

Given $P(A' \cap B') = \frac{1}{2}$:

(ii) $P(B)$

We know the formula for conditional probability: $P(B|A) = \frac{P(A \cap B)}{P(A)}$.

Substitute the known values:

Find a common denominator (12):

8b) A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addresses the envelopes, at random, one each to A, B, C, D and E. Find the probability that
The total number of ways to address the 5 envelopes is $5! = 120$.

(i) the letter to A is in the correct envelope and the letter to B is in an incorrect envelope.

The letter to A is in the correct envelope (addressed to A). There is only 1 way for this.
Now, 4 letters ($L_B, L_C, L_D, L_E$) remain for 4 envelopes ($E_B, E_C, E_D, E_E$).

The letter to B is in an incorrect envelope (not addressed to B).
Total permutations for the remaining 4 letters is $4! = 24$.
Number of permutations where $L_B$ is in $E_B$: If $L_B$ is in $E_B$, then the remaining 3 letters ($L_C, L_D, L_E$) can be arranged in the remaining 3 envelopes ($E_C, E_D, E_E$) in $3! = 6$ ways.
So, the number of ways where $L_B$ is NOT in $E_B$ is $4! - 3! = 24 - 6 = 18$.

The number of favorable outcomes is 18. The total number of outcomes is $5! = 120$.

(ii) the letter to A is in the correct envelope, given that the letter to B is in an incorrect envelope.
Let $C_A$ be the event "letter to A is in the correct envelope".
Let $I_B$ be the event "letter to B is in an incorrect envelope".
We need to find $P(C_A | I_B) = \frac{P(C_A \cap I_B)}{P(I_B)}$.

This is the probability calculated in part (i): $P(C_A \cap I_B) = \frac{18}{120}$.

The total number of ways to address 5 envelopes is $5! = 120$.
Number of ways where $L_B$ is in $E_B$: If $L_B$ is in $E_B$, the remaining 4 letters can be arranged in $4! = 24$ ways.
So, the number of ways where $L_B$ is NOT in $E_B$ is $5! - 4! = 120 - 24 = 96$.

Simplify the fraction: