Catching Errors While Proofreading a Document

A document has n typos, and there are two proofreaders working independently on proofreading the document. Proofreader A catches typos with probability $p_{1}$ and misses them with probability $q_{1} = 1 - p_{1}$. Proofreader B catches typos with probability $ p_{2}$ and misses them with probability, you guessed it, $q_{2} = 1 - p_{2}$. Let $X_{1}$ be the number of typos caught by Proofreader A, and $X_{2}$ be the number of typos caught by Proofreader B, and $X$ be the number of typos caught by at least one of them.

(a) What is the distribution of X?

Solution: The easiest way to think about this is to think of the case where neither of them catch a typo. The probability of both of them missing a typo, (each Proofreader's miss probability is independent of the other's), is gotten through multiplication, $q_{1}q_{2}$. So, the probability of at least one of them catching that typo is $1 - q_{1}q_{2}$. Thus, the distribution is
$X\sim Bin(n,1-q_{1}q_{2})$

(b) Assume $p_{1} = p_{2}$, what is the conditional distribution of  $X_{1}$ given that $X_{1}+X_{2} = t$?

Solution: Using Bayes' Rule, we have:

$P(X_{1}=k|T=t) = \frac{P(T=t|X_{1}=k)P(X_{1}=k)}{P(T=t)}$


with $T\sim Bin(2n,p)$

$P(X_{1}=k|T=t) = \frac{\binom{n}{t-k}p^{t-k}q^{n-t+k}\binom{n}{k}p^{k}q^{n-k}}{\binom{2n}{t}p^{t}q^{2n-t}} = \frac{\binom{n}{t-k}\binom{n}{k}}{\binom{2n}{t}}$

$X_{1}\sim HGeom(n,n,t)$

Notes: $P(T=t|X_{1}=k)$ considers the probability of T = t given $X_{1}=k$, which means we only have to consider how $X_{2}$ affects the calculation. t = total catches by Proofreader A + total catches by Proofreader B. If A catches k typos, then B catches t - k typos.

Welcome to the Party!

People arrive at a party one at a time. Those who are still waiting for all the other guests to arrive amuse themselves by seeing if they share their birthday with anyone else. Up until the Xth guest arrives, no matches have been made. Let X be the guest who shares a birthday with one of the guests who arrived before. Find P(X = 3 or X = 4).

Solution: First we'd like to find the PMF and then sum the series of terms with X = 3 and X = 4. It's easier to start by considering the first k - 1 guests and use the fact that no one shares a birthday in that group. We could write out an expression that demonstrates this fact easily enough. Let the kth guest be the one who does share a birthday with one of the k - 1 guests.

$P(X\leq k-1) = 365*\left ( \frac{1}{365} \right )*364*\left ( \frac{1}{365} \right )...\left ( 365-k+2 \right )*\left ( \frac{1}{365} \right )$

We have k - 1 guests, so we have k-1 factors in the numerator and k-1 factors in the denominator of the following simplification of the above equation:

$P(X\leq k-1) = \frac{365*364...(365-k+2)}{365^{k-1}}$

When the kth guest shows up, we expect to get a match. There are k-1 ways to get a match with this element, and the birthday probability is given, as usual, as 1/365. So, now we have:

$P(X\geq k) = \frac{365*364...(365-k+2)}{365^{k-1}}*\left ( \frac{k-1}{365} \right )$

P(X = 3 or X = 4) =

$\frac{365*364...(365-3+2)}{365^{3-1}}*\left ( \frac{3-1}{365} \right ) + \frac{365*364...(365-4+2)}{365^{4-1}}*\left ( \frac{4-1}{365} \right ) \approx 0.0136$

It's worth trying this exercise with an application of Bayes' Rule...

$P(X=k(match)|X< k(no-match)) = \frac{P(X<k (no-match)|X=k (match))P(X=k (match))}{P(X< k (no-match))}$

Counting Chickens After They Hatch

Given n eggs, let the probability of an egg hatching be p, each hatching event being independent of the others. The probability of a chick surviving after hatching we'll call r, which is independent of the survival of other hatchlings. Let H be the number of eggs that hatch and C be the number of chicks that survive. What are the distributions of H and C?

Solution: H ~ Bin(n, p), which is pretty straightforward. We're given that post-hatch survival is C, with a probability independent of other hatched chicks. Chickens that end up surviving, according to this set up, go through two stages, hatching and surviving. We're also given that the probability of surviving after hatching is r, which is independent of p, the probability of hatching. So, the probability of survival for a single chick is $p * r$. This gives the distribution of C.

$H\sim Bin(n, p)$
$C\sim Bin(n, rp)$

Choose the Coin, Then Flip it n Times

Say there are two coins to choose from, and you choose one of them randomly with each one just as likely to be chosen as the other. Then you flip that coin n times with a flip ending up as heads being a success. Say the probability of coin 1 landing heads is $p_{1}$ and the probability of coin 2 landing heads is $p_{2}$. Let X be the number of times a coin lands heads.

(a) What's the Probability Mass Function for this distribution?

We can look at the beginnings of a tree diagram and get an idea of how to proceed.

If we choose Coin 1, we proceed along a path that has nothing to do with Coin 2, and the same logic holds if we choose Coin 2. So, we're looking at combining two cases. 

Case 1:

$P_{1}(X=k) = \frac{1}{2}\binom{n}{k}p_{1}^{k}(1-p_{1})^{n-k}$

Case 2:

$P_{2}(X=k) = \frac{1}{2}\binom{n}{k}p_{2}^{k}(1-p_{2})^{n-k}$

Combine the two to get:

$P(X=k) = \frac{1}{2}\binom{n}{k}p_{1}^{k}(1-p_{1})^{n-k} + \frac{1}{2}\binom{n}{k}p_{2}^{k}(1-p_{2})^{n-k}$

Verify this result using the Law of Total Probability (LOTP):

$P(X=k) = P\left ( X=k | C_{1} \right )P(C_{1}) + P\left ( X=k | C_{2} \right )P(C_{2})$

Where $C_{1}$ is the event Coin 1 is chosen and similarly for Coin 2. 

$P\left ( C_{1} \right )= P\left ( C_{2} \right )= \frac{1}{2}$

Is This a Valid Probability Mass Function?

The first-digit law, or Benford's Law, states, according to Wikipedia, that in "many naturally occurring collections of numbers the leading significant digit is likely to be small. For example, in sets that obey the law, the number 1 appears as the most significant digit about 30% of the time, while 9 appears as the most significant digit about 5% of the time."

$P(D=j)= log_{10}\left ( \frac{j+1}{j} \right ), for j\in \left \{ 1,2,3,...,9 \right \}$

with D being the first digit. Is this a valid PMF?

$ \sum_{j=1}^{9}log_{10}\left ( \frac{j+1}{j} \right )=  \sum_{j=1}^{9}log_{10}\left ( j+1 \right )-log_{10}\left ( j \right )$

= 1.

Also, the sum is non-negative, so yes, it is a valid PMF.

Win, Lose or Draw: 7 Games of Chess

1.) Two chess players, X and Y, play a 7-game series of chess. For each game, there are three possible outcomes for a player, win, lose or draw. A win nets the victor one point and the loser zero points. In the case of a draw, each player is awarded a half-point. How many ways are there for one of the players, say, player X, to get an overall result of 3 wins, 2 draws and 2 losses?

Solution: A simple calculation here... We multiply the number of ways a player can win 3 out of the 7 games, the number of ways that player can get a draw or a loss in 4 of the seven games, and the number of ways that player can get a draw or a loss in 2 out of the 7 games. The factors are:
$\binom{7}{3}, \binom{4}{2}, \binom{2}{2}$

The answer is $\binom{7}{3}* \binom{4}{2}* \binom{2}{2}$ = 210

2. How many possible outcomes for the games are there with X ending up with 4 points and Y ending up with 3 points?

Solution: One way is to have X win 4 out of the 7 games and lose 3: $\binom{7}{4}*\binom{3}{3}$. Another way is to have X win 3, draw 2, and lose 2: $\binom{7}{3}* \binom{4}{2}* \binom{2}{2}$. Another way is to have X win 2, draw 4 and lose 1: $\binom{7}{2}* \binom{5}{4}* \binom{1}{1}$. And, finally, another way is to have X win 1 and draw 6: $\binom{7}{1}* \binom{6}{6}$

Answer = $ \binom{7}{4}*\binom{3}{3} + \binom{7}{3}* \binom{4}{2}* \binom{2}{2} + \binom{7}{2}* \binom{5}{4}* \binom{1}{1} + \binom{7}{1}* \binom{6}{6}$
= 35 + 210 + 105 + 7 = 357.

3. How many possible outcomes for the games are there where one player ends up with 4 points, the other with 3 points, and the full 7 games are played?

Solution: The 7th game must be played, which means that for one player, say, X to end up with 4 points at the end, he or she cannot lose that 7th game. Therefore, the only possibilities for X to end up with 4 points are either a full-point awarded in the seventh game, or only a half point. If the full point is awarded in the 7th game, that means X must get 3 points in the first 6 games. If only a half-point is awarded in the 7th game, then X must get 3.5 points in the first 6 games.

Let scenario 1 be the scenario where X gets a full point in the 7th game. This gives us for the first 6 games:

$\binom{6}{3}* \binom{3}{3}$ = 20 (3 wins and 3 losses)
$\binom{6}{2}* \binom{4}{2}* \binom{2}{2}$  = 90 (2 wins, 2 draws, and 2 losses)
$\binom{6}{1}* \binom{5}{4}* \binom{1}{1}$ = 30 (1 win, 4 draws, and 1 loss)
$\binom{6}{0}* \binom{6}{6}$ = 1 (0 wins and 6 draws)

Let scenario 2 be the one where X gets only a half point in the 7th game. This gives us for the first 6 games:

$\binom{6}{3}* \binom{3}{1}*\binom{2}{2}$ = 60 (3 wins, 1 draw, and 2 losses)
$\binom{6}{2}* \binom{4}{3}*\binom{1}{1}$ = 60 (2 wins, 3 draws, and 1 loss)
$\binom{6}{1}* \binom{5}{5}$ = 6 (1 win and 5 draws)

Answer = 20 + 90 + 30 + 1 + 60 + 60 + 6 = 267

Phone Numbers as Possible Combinations of 7 digits: Probability

1.) If the first digit isn't allowed to be a 0 or a 1, how many 7-digit phone numbers are possible?

Solution: For each digit in the phone number, there are 10 possibilities (0 - 9) if there are no restrictions. In this case the first number isn't allowed to be 0 or 1. For phone-number digits 2 - 7, there are $10^{6}$ possibilities: 10 * 10 * 10 * 10 * 10 * 10. Multiply this by the number of possibilities for the first digit (only 8 since two digits, 1 and 0, are excluded), and you get your answer:

No. of possibilities for the first digit * No. of possibilities for the remaining 6 digits

= $8*10^{6}$ = 8,000,000 possible phone numbers.

2.) If the above condition stays in place, along with the added condition that the first three digits of the phone number cannot be 911, how many possible phone numbers are there?

Solution: If the probability of an event A occurring is P(A), then the probability of event A not occurring is 1 - P(A). Consider the case where the event A is the choosing of the first three digits of the phone number to be 911:

The probability space of the first three digits is $8*10*10$ if we consider only the condition of part 1. The probability of the first three digits being 911 is $\frac{1}{8}*\frac{1}{10}*{\frac{1}{10}} = \frac{1}{800}$

So, the probability of the first three digits not being 911 is $ 1 -\frac{1}{800} = \frac{799}{800}$

Finally, the number of possible phone numbers with the first digit not being a 0 or a 1, and the first 3 digits not being 911, is
$799 * 10^{4} $ = 7,990,000.

(799 is the number of possibilities for the first 3 digits, and $10^{4}$ is the number of possibilities for the last 4 digits.)