Let $X$ and $Y$ be geometric random variables with parameter $p$, with $0 \leq p \leq 1$. Assume that $X$ and $Y$ are independent.

Let $n$ be an integer greater than $1$. Let $k$ be a natural number with $k\leq n$. Then prove the formula
\[P(X=k \mid X + Y = n) = \frac{1}{n-1}.\]

Hint.

If this problem is a bit abstract for you, then you might want to try the following problem, which is more concrete.

Probability that Alice Tossed a Coin Three Times If Alice and Bob Tossed Totally 7 Times

Solution.

The definition of a conditional probability yields
\[P(X=k \mid X + Y = n) = \frac{P\left((X=k) \cap (X+Y=n)\right)}{P(X+Y=n)} \tag{*}\] We first compute the numerator of the last expression. We have
\begin{align*}
&P\left((X=k) \cap (X+Y=n)\right)\\
&= P\left((X=k) \cap (Y=n-X)\right)\\
&= P\left((X=k) \cap (Y=n-k)\right)\\
&= P(X=k) \cdot P(Y=n-k).
\end{align*}
Here, the last step follows because $X$ and $Y$ are independent.

Now, as $X$ and $Y$ are geometric random variables with parameter $p$, we see that by putting $q=1-p$
\begin{align*}
P(X=k) &= q^{k-1}p \\
P(Y= n – k) &= q^{n-k-1}p.
\end{align*}
Hence, the numerator becomes
\begin{align*}
&P\left((X=k) \cap (X+Y=n)\right)\\
&= P(X=k) \cdot P(Y=n-k)\\
&= q^{k-1}p \cdot q^{n-k-1}p\\
&= q^{n-2}p^2.
\end{align*}

Next, we compute the denominator of (*). Applying the law of total probability, we get
\begin{align*}
P(X+Y=n) &= \sum_{k=1}^{n-1} P(X=k) P(X+Y=n \mid X=k). \tag{**}
\end{align*}
The probability $P(X+Y=n \mid X=k)$ can be computed as follows.
\begin{align*}
&P(X+Y=n \mid X=k)\\
&= P(Y = n-X \mid X=k)\\
&= P(Y=n-k \mid X=k)\\
&= P(Y=n-k).
\end{align*}
The last step follows because $X$ and $Y$ are independent. Then the equation (**) becomes
\begin{align*}
P(X+Y=n) &= \sum_{k=1}^{n-1} P(X=k) \cdot P(Y=n-k) \\ &= \sum_{k=1}^{n-1} q^{k-1}p \cdot q^{n-k-1}p\\
&= \sum_{k=1}^{n-1} q^{n-2} p^2\\
&= (n-1)q^{n-2}p^2.
\end{align*}

Combining the computations of the numerator and the denominator, we obtain the desired probability from (*).
\begin{align*}
P(X=k \mid X + Y = n) &= \frac{P\left((X=k) \cap (X+Y=n)\right)}{P(X+Y=n)}\\
&= \frac{q^{n-2}p^2}{(n-1)q^{n-2}p^2}\\
&= \frac{1}{n-1}.
\end{align*}
This completes the proof of the formula
\[P(X=k \mid X + Y = n) = \frac{1}{n-1}.\]

Application

As an application of this problem, solve the next problem.

Its solution is available in the post Probability that Alice Tossed a Coin Three Times If Alice and Bob Tossed Totally 7 Times.

The post Conditional Probability When the Sum of Two Geometric Random Variables Are Known first appeared on Problems in Mathematics.

We have a stick of a unit length. Two points on the stick will be selected randomly (uniformly along the length of the stick) and independently. Then we break the stick at these two points so that we get three pieces of the stick. What is the probability that these three pieces form a triangle?

Solution.

Let us call the left end of the stick the origin. Let $X$ be the length from the origin to the first selected point. Let $Y$ be the length from the origin to the second selected point.

Note that when $X=Y$, we have only two pieces and they cannot form a triangle, so we assume $X \neq Y$.
We first assume that $X < Y$. Then after breaking the stick, we have three pieces of length $X$, $Y-X$, and $1-Y$, respectively.

Now, let us consider a condition so that three segments of length $a$, $b$, and $c$ form a triangle in general. It is necessary and sufficient that the sum of the lengths of any two segments is bigger than the length of the rest to form a triangle.
Namely, we must satisfy all of the following inequalities.
\begin{align*}
a + b &> c\\
b + c &> a\\
c + a &> b.
\end{align*}

Applying this condition to the current situation, we must have
\begin{align*}
X + (Y-X) &> 1-Y\\
(Y-X) + (1-Y) & > X\\
(1- Y) + X &> Y – X.
\end{align*}
Simplifying these, we obtain
\begin{align*}
Y &> \frac{1}{2}\\
X &< \frac{1}{2}\\ Y &< X + \frac{1}{2}. \end{align*} The region in the unit square satisfying these inequality is depicted in the figure below (orange region).

Note that the joint density function of $X$ and $Y$ are uniformly distributed on the unit square. Thus, the probability that three pieces form a triangle under the assumption $X < Y$ is given by the area of the orange triangular region, which is $1/8$. Note that by symmetry, the case $X > Y$ gives the same probability. Hence, the desired probability is
\begin{align*}
&P(\text{form a triangle}) \\
&= P(\text{form a triangle} \mid X < Y) + P(\text{form a triangle} \mid X > Y)\\
&\frac{1}{8}+\frac{1}{8} = \frac{1}{4}.
\end{align*}

Alternatively, we can find the probability for the case $X > Y$ as follows. By symmetry argument, we simply need to exchange $X$ and $Y$ and obtain the conditions
\begin{align*}
X &> \frac{1}{2}\\
Y &< \frac{1}{2}\\ X &< Y + \frac{1}{2}. \end{align*} The blue region below shows the area surrounded by these inequalities. The area of this blue region is $1/8$. Thus, the total area is $1/8 + 1/8 = 1/4$ as before.

The post Probability that Three Pieces Form a Triangle first appeared on Problems in Mathematics.

Alice tossed a fair coin until a head occurred. Then Bob tossed the coin until a head occurred. Suppose that the total number of tosses for Alice and Bob was $7$.

Assuming that each toss is independent of each other, what is the probability that Alice tossed the coin exactly three times?

Hint.

Let $X$ be the number of Alice’s tosses until she got a head.

Then $X$ is a geometric random variable with parameter $p=1/2$.
Recall that a geometric random variable can be interpreted as the number of trials until getting the first success and each trial is successful with probability $p$ and fails with probability $q=1-p$.

Note that
\[P(X=k)= q^{k-1} p = (1-p)^{k-1} p.\]

Solution.

Let $X$ be the number of Alice’s tosses until she got a head. Also, let $Y$ be the number of Bob’s tosses until he got a head. As each toss is an independent event and the coin is fair, the random variables $X$ and $Y$ are geometric with parameter $p=1/2$ (the probability of success(head) is $p=1/2$ and the probability of failure(tail) is $1-p=1/2$).

This question asks to find the conditional probability
\[P(X=3 \mid X+Y=7) = \frac{P\left((X=3) \cap (X+Y=7) \right)}{P(X+Y=7)}.\]

Let us first compute the numerator as follows.
\begin{align*}
P\left((X=3) \cap (X+Y=7) \right) &= P\left((X=3) \cap (Y=7-X) \right) \\
&= P\left((X=3) \cap (Y=4) \right) \\
&= P(X=3) \cdot P(Y=4)
\end{align*}
since $X$ and $Y$ are independent.

Now, as $X$ and $Y$ follow the geometric distribution with parameter $p$, we have
\begin{align*}
P(X=3) &= \left(\frac{1}{2}\right)^2 \frac{1}{2} = \frac{1}{2^3}\\
P(Y=4) &= \left(\frac{1}{2}\right)^3 \frac{1}{2} = \frac{1}{2^4}.
\end{align*}
The product of these gives the numerator $\frac{1}{2^7}$.

Next, let us calculate the denominator $P(X+Y=7)$.
The law of total probability yields
\begin{align*}
P(X+Y=7) = \sum_{k=1}^6 P(X=k) P(X+Y = 7 \mid X=k).
\end{align*}

As before, since $X$ is a geometric random variable with parameter $p=1/2$, we have
\[P(X=k) = \left(\frac{1}{2}\right)^{k-1} \frac{1}{2} = \frac{1}{2^k}.\]

The second factor $P(X+Y \mid X= k)$ can be computed as follows.
\begin{align*}
&P(X+Y = 7 \mid X= k)\\
&= P(Y = 7 – X \mid X = k)\\
&= P(Y = 7 – k \mid X = k)\\
&= P(Y=7 – k)\\
&= \left(\frac{1}{2}\right)^{(7-k)-1} \frac{1}{2} = \frac{1}{2^{7-k}}
\end{align*}
Here, the third equality follows because $X$ and $Y$ are independent.

It follows that the denominator becomes
\begin{align*}
P(X+Y=7) &= \sum_{k=1}^6 \frac{1}{2^k} \cdot \frac{1}{2^{7-k}}\\
&= \sum_{k=1}^6 \frac{1}{2^7}\\
&= \frac{6}{2^7}.
\end{align*}

Combining these computations, it follows that the desired conditional probability is
\begin{align*}
P(X=3 \mid X+Y=7) = \frac{1/2^7}{6/2^7}=\frac{1}{6}.
\end{align*}

Abstraction

If you solved this problem, then try the following problem as well. This is more abstract than the current problem but the idea is the same.

Its solution can be found in the post: Conditional Probability When the Sum of Two Geometric Random Variables Are Known

The post Probability that Alice Tossed a Coin Three Times If Alice and Bob Tossed Totally 7 Times first appeared on Problems in Mathematics.

A researcher conducted the following experiment. Students were grouped into two groups. The students in the first group had more than 6 hours of sleep and took a math exam. The students in the second group had less than 6 hours of sleep and took the same math exam.

The pass rate of the first group was twice as big as the second group. Suppose that $60\%$ of the students were in the first group. What is the probability that a randomly selected student belongs to the first group if the student passed the exam?

Solution.

Let $E$ be the event that a student passes the exam.
Let $G_i$ be the event that a student belongs to the group $i$ for $i= 1, 2$. Then the desired probability is $P(G_1 \mid E)$. ($G_1$ is the first group and $G_2$ is the second group.)

Using these notation, the pass rate of the first group is expressed as $P(E \mid G_1)$. Similarly, the pass rate of the second group is $P(E \mid G_2)$. By assumption, the pass rate of the first group is twice as big as the second group. Hence, we have
\[P(E \mid G_1) = 2\ P(E \mid G_2).\]

The required probability that a randomly selected student belongs to the first group given that the student passes the exam is expressed as $P(G_1 \mid E)$.

Using the above equality and Bayes’ rule, we get
\begin{align*}
&P(G_1 \mid E)\\[6pt] &= \frac{P(G_1) P(E \mid G_1)}{P(E)} & \text{(by Bayes’ rule)}\\[6pt] &= \frac{P(G_1)P(E \mid G_1)}{P(E \mid G_1)P(G_1) + P(E \mid G_2)P(G_2)} & \text{(rule of total probability)}\\[6pt] &= \frac{P(G_1)\cdot 2P(E \mid G_2)}{2P(E \mid G_2)P(G_1) + P(E \mid G_2)P(G_2)} \\[6pt] &= \frac{P(G_1)\cdot 2}{2P(G_1) + P(G_2)}\\[6pt] &= \frac{0.6 \cdot 2}{2\cdot 0.6 + 0.4}\\[6pt] &= \frac{3}{4}.
\end{align*}

Therefore, the required probability is $P(G_1 \mid E) = \frac{3}{4}$.

The post Find the Conditional Probability About Math Exam Experiment first appeared on Problems in Mathematics.

There are three coins in a box. The first coin is two-headed. The second one is a fair coin. The third one is a biased coin that comes up heads $75\%$ of the time. When one of the three coins was picked at random from the box and tossed, it landed heads.

What is the probability that the selected coin was the two-headed coin?

Hint.

Use Bayes’ theorem (Bayes’ rule).

Solution.

Let $E_i$ be the event of the $i$-th coin being picked for $i = 1, 2, 3$. Let $F$ be the event that a coin lands heads.

The required probability can be calculated using Bayes’ rule as follows:
\[P(E_1 \mid F) = \frac{P(F \mid E_1) \cdot P(E_1)}{P(F)}.\] When the two-headed coin is picked, it always lands heads. Thus, we have the conditional probability $P(F \mid E_1) = 1$. The probability that the two-headed coin is selected out of the box is $P(E_1)=1/3$.

The probability in the denominator is calculated using the total probability theorem as follows:
\begin{align*}
P(F) &= \sum_{i=1}^3 P(F \mid E_i) P(E_i) \\[6pt] &= \frac{1}{3}\left(1+\frac{1}{2} + \frac{3}{4} \right)\\[6pt] &= \frac{3}{4}.
\end{align*}

It follows by the formula above that the required probability is
\[P(E_1 \mid F) = \frac{1\cdot \frac{1}{3}}{\frac{3}{4}} = \frac{4}{9}.\] Click here if solved 14

The post What is the Probability that Selected Coin was Two-Headed? first appeared on Problems in Mathematics.

A certain model of smartphone is manufactured by three factories A, B, and C. Factories A, B, and C produce $60\%$, $25\%$, and $15\%$ of the smartphones, respectively. Suppose that their defective rates are $5\%$, $2\%$, and $7\%$, respectively.

If a smartphone of this model is found out to be detective, what is the probability that this smartphone was manufactured in factory C?

Hint.

Use the Bayes’ theorem.

Solution.

Let $E$ be the event that a smartphone of this model is defective. Let $F_A$ be the event that a smartphone is manufactured by factory A. Similarly for $F_B$ and $F_C$.

By Bayes’s rule, we have
\[P(F_C \mid E) = \frac{P(F_C) \cdot P(E \mid F_C)}{P(E)}.\]

Now, we compute the probabilities on the right hand side.

In the post Overall Fraction of Defective Smartphones of Three Factories, we calculated that
\begin{align*}
P(E) &= P(F_A)\cdot P(E \mid F_A) + P(F_B)\cdot P(E \mid F_B) + P(F_C)\cdot P(E \mid F_C)\\
&= 0.0455.
\end{align*}
(See the post for details.)

Factory C produces $15\%$ of the smartphones, thus $P(F_C)=0.15$.
Also, the defective rate for Factory C is $7\%$. Hence $P(E \mid F_C) = 0.07$.

Inserting these values into the formula above, we get
\[P(F_C \mid E) = \frac{0.15 \cdot 0.07}{0.0455} \approx 0.2308.\] Click here if solved 8

The post If a Smartphone is Defective, Which Factory Made It? first appeared on Problems in Mathematics.

Two fair coins are tossed. Given that at least one of them lands heads, what is the conditional probability that the first coin lands heads?

We give two proofs. The first one uses Bays’ theorem and the second one simply uses the definition of the conditional probability.

Solution 1.

Let $E$ be the event that the first coin lands heads. Let $F$ be the event that at least one of two coins lands heads.
By Bayes’ rule, the required probability can be calculated by the formula:
\[P(E \mid F) = \frac{P(E) \cdot P(F \mid E)}{P(F)}.\] We know $P(E)=1/2$. When the first coin lands heads, then of course at least one of two coins lands heads. So, we have $P(F \mid E) = 1$.

Since $F = \{\text{hh}, \text{ht}, \text{th}\}$, we see that $P(F) = 3/4$.
Plugging these values into the formula, we obtain
\[P(E \mid F) = \frac{\frac{1}{2}\cdot 1}{\frac{3}{4}} = \frac{2}{3}.\]

Solution 2.

Let $E$ be the event that the first coin lands heads. Let $F$ be the event that at least one of two coins lands heads.
Then we have $F = \{\text{hh}, \text{ht}, \text{th}\}$.
Also, we have
\[E \cap F = \{\text{hh}, \text{ht}\}.\] Thus, the required probability is given by
\begin{align*}
P(E \mid F) &= \frac{|E \cap F|}{|F|}\\
&= \frac{2}{3}.
\end{align*}

The post If At Least One of Two Coins Lands Heads, What is the Conditional Probability that the First Coin Lands Heads? first appeared on Problems in Mathematics.

Let $C$ be the event that a randomly chosen person has lung cancer. Let $S$ be the event of a person being a smoker.
Suppose that 10% of the population has lung cancer and 20% of the population are smokers. Also, suppose that we know that 70% of all people who have lung cancer are smokers.

Then determine the probability of a person having lung cancer given that the person is a smoker.

Solution.

Hint (Bayes’s rule)

Let $E$ and $F$ be events.

Let $P(E \mid F)$ be the probability that $E$ occurs given $F$ occurs. This is called a conditional probability of $E$ given $F$.

Suppose that we know $P(E), P(F)$ and $P(E \mid F)$. Then $P(F \mid E)$ can be computed by Bayes’ theorem (alternatively Bayes’ rule):
\[ P(E \mid F) = \frac{P(E) \cdot P(F \mid E)}{P(F)}.\]

Solution

Given information can be formulated as
\[P(C) = 0.1, P(S) = 0.2, \text{ and } P(S \mid C) = 0.7.\]

The required probability is $P(C \mid S)$. Using Bayes’ rule, we can compute it as follows.
\begin{align*}
P(C \mid S) &= \frac{P(C) \cdot P(S \mid C)}{P(S)}\\[6pt] &= \frac{(0.1)(0.7)}{0.2}\\[6pt] &= 0.35
\end{align*}

Remark

The data given here is artificial for educational purpose and is not based on a scientific fact.

The post Probability of Having Lung Cancer For Smokers first appeared on Problems in Mathematics.

A card is chosen randomly from a deck of the standard 52 playing cards.

Let $E$ be the event that the selected card is a king and let $F$ be the event that it is a heart.

Prove or disprove that the events $E$ and $F$ are independent.

Definition of Independence

Events $E$ and $F$ are said to be independent if
\[P(E \cap F) = P(E) \cdot P(F).\]

Intuitively, this means that an occurrence of one does not change the probability that the other occurs. Mathematically, this can be seen as follows.

If $P(E) \neq 0$, then independency implies
\[P(F \mid E) = \frac{P(F \cap E)}{P(E)} = \frac{P(F)P(E)}{P(E)} = P(F).\]

Solution.

We prove that the events $E$ and $F$ are independent.
First, we have $P(E \cap F) = 1/52$ because there is only one King of hearts card in the deck of 52 cards.

Since there are four kings, we have $P(E) = 4/52$. As there are $13$ heart cards, we have $P(F) = 13/52$.
Thus, we see that
\[P(E)P(F) = \frac{4}{52} \cdot \frac{13}{52} = \frac{1}{52} = P(E \cap F).\] This implies that the events $E$ and $F$ are independent.

The post Independent Events of Playing Cards first appeared on Problems in Mathematics.

A jewelry company requires for its products to pass three tests before they are sold at stores. For gold rings, 90 % passes the first test, 85 % passes the second test, and 80 % passes the third test. If a product fails any test, the product is thrown away and it will not take the subsequent tests. If a gold ring failed to pass one of the tests, what is the probability that it failed the second test?

Solution.

Let $F$ be the event that a gold ring fails one of the three tests. Let $F_2$ be the event that it fails the second test. Then what we need to compute is the conditional probability
\[P(F_2 \mid F) = \frac{P(F_2 \cap F)}{P(F)}.\]

The numerator is
\[P(F_2 \cap F) = P(F_2) = 0.9 \cdot 0.15.\] (A gold ring passes the first test with probability $0.9$ and fails the second test with probability $1-0.85=0.15$.)

The complement $F^c$ of $F$ is the event that a gold ring passes all the tests. Thus
\[P(F) = 1- P(F^c) = 1 – 0.9 \cdot 0.85 \cdot 0.8.\] It follows that the desired probability is
\begin{align*}
P(F_2 \mid F) &= \frac{0.9 \cdot 0.15}{1 – 0.9 \cdot 0.85 \cdot 0.8}
= \frac{135}{388} \approx 0.348
\end{align*}

Therefore, given that a gold ring failed to pass one of the tests, the probability that it failed the second test is about 34.8 %.

The post Jewelry Company Quality Test Failure Probability first appeared on Problems in Mathematics.