Alice and Bob play some game against each other. The probability that Alice wins one game is $p$. Assume that each game is independent.

If Alice wins $n$ games before Bob wins $m$ games, then Alice becomes the champion of the game. What is the probability that Alice becomes the champion.

Solution.

If Alice won at least $n$ games in the first $n+m-1$ games, then Bob won at most $m-1$. On the other hand, if Alice won at most $n-1$ games in the first $n+m-1$, then Bob won at least $m$ games. Hence, it is necessary and sufficient for Alice to be the champion that Alice wins at least $n$ games in the first $n+m-1$ games.

(Remark that even if Alice won $n$ games before finishing $n+m-1$ games, we may assume that they continue playing until $n+m-1$ games.)

The probability that Alice wins exactly $k$ games in the first $n+m-1$ games is given by
\[{n+m-1 \choose k} p^k(1-p)^{n+m-k-1}\] since each game is independent.
Therefore, the probability that Alice becomes the champion is obtained by summing this probability over $k=n, n+1, \dots, n+m-1$.

Hence, we get
\[P(\text{Alice becomes the champion}) = \sum_{k=n}^{n+m-1}{n+m-1 \choose k} p^k(1-p)^{n+m-k-1}.\] Click here if solved 4

The post Probability that Alice Wins n Games Before Bob Wins m Games first appeared on Problems in Mathematics.

Consider an infinite series of events of rolling a fair six-sided die. Assume that each event is independent of each other. For each of the below, determine its probability.

(1) At least one die lands on the face 5 in the first $n$ rolls.
(2) Exactly $k$ dice land on the face 5 in the first $n \geq k$ rolls.
(3) Every die roll results in the face 5.

Solution.

Solution of (1)

Let $E_i$ be the $i$th die roll event that lands on the face 5. Then we have the probability $P(E_i) = 1/6 =: p$. Then the desired probability is obtained as follows.
\begin{align*}
&P(\text{at least one 5 in the first $n$ trials}) \\
&= 1 – P(\text{no 5 in the first $n$ trials})\\
&= 1 – P(E_1^c E_2^c \cdots E_n^c)\\
&= 1 – P(E_1^c) P(E_2^c) \cdots P(E_n^c) && \text{ by independence}\\
&= 1 – \left(1-P(E_1)\right) \left(1-P(E_2)\right) \cdots \left(1-P(E_n)\right)\\
&= 1 – (1-p)^n\\
&= 1 – (5/6)^n
\end{align*}

Solution of (2)

Suppose that we have exactly $k$ dice land on the face 5 and the rest of $n-k$ trials land other than 5. The probability of such an event is given by $p^k(1-p)^{n-k}$ as each trial is independent of each other. Now, there are totally ${n \choose k}$ such sequences of events. Thus, the probability that exactly $k$ dice land on the face 5 in the fist $n$ rolls is
\[{n \choose k}p^k(1-p)^{n-k} = {n \choose k}\left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{n-k} = {n \choose k} \frac{5^{n-k}}{6^n}.\]

Solution of (3)

Let $F_i$ be the event that all die rolls land the face 5 for the first $i$th trials. Thus, $F_i = E_1 E_2 \cdots E_i$. Note that the sequence $\{F_i\}$ is increasing, that is, we have
\[F_1 \subset F_2 \subset \cdots \subset F_n \subset \cdots.\]

Now, the desired probability can be expressed as $P(\lim_{n\to \infty} F_n)$. By the result of Problem “Interchangeability of Limits and Probability of Increasing or Decreasing Sequence of Events“, we obtain
\begin{align*}
&P\left(\lim_{n\to \infty} F_n \right)\\
&= \lim_{n \to \infty} P(F_n)\\
&= \lim_{n \to \infty} P(E_1 E_2 \dots E_n)\\
&= \lim_{n \to \infty} P(E_1) P(E_2) \cdots P(E_n) && \text{ by independence}\\
&= \lim_{n \to \infty} p^n \\
&= \lim_{n \to \infty} (1/6)^n\\
&= 0.
\end{align*}
Therefore the probability that all die rolls result in the face 5 is zero.

The post Probabilities of An Infinite Sequence of Die Rolling first appeared on Problems in Mathematics.

Let $E$ and $F$ be independent events. Let $F^c$ be the complement of $F$.

Prove that $E$ and $F^c$ are independent as well.

Solution.

Note that $E\cap F$ and $E \cap F^c$ are disjoint and $E = (E \cap F) \cup (E \cap F^c)$. It follows that
\[P(E) = P(E \cap F) + P(E \cap F^c).\] As $E$ and $F$ are independent, we know that
\[P(E \cap F) = P(E)\cdot P(F)\] Combining these two equalities, we get
\begin{align*}
P(E \cap F^c) &= P(E) – P(E \cap F)\\
&= P(E) – P(E) \cdot P(F)\\
&= P(E)(1-P(F)).
\end{align*}

Since $P(F^c) = 1 – P(F)$, we obtain the equality
\[P(E \cap F^c) = P(E)\cdot P(F^c),\] which implies that $E$ and $F^c$ are independent.

Remark

We just proved that when $E$ and $F$ are independent events, then $E$ and the complement $F^c$ are independent.
Now, we apply this statement to the independent events $E$ and $F^c$. Then we see that the complements $E^c$ and $F^c$ are independent.

In conclusion, if two events are independent, then their complements are also independent.

The post Complement of Independent Events are Independent first appeared on Problems in Mathematics.

Suppose that three fair coins are tossed. Let $H_1$ be the event that the first coin lands heads and let $H_2$ be the event that the second coin lands heads. Also, let $E$ be the event that exactly two coins lands heads in a row.

For each pair of these events, determine whether they are independent or not.

Definition of Independence

Solution.

The post Independent and Dependent Events of Three Coins Tossing 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.