coin toss – Problems in Mathematics

What is the Probability that Selected Coin was Two-Headed?

Yu — Wed, 30 Oct 2019 06:37:27 +0000

Problem 739

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
\[P(E_1 \mid F) = \frac{1\cdot \frac{1}{3}}{\frac{3}{4}} = \frac{4}{9}.\]

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

If At Least One of Two Coins Lands Heads, What is the Conditional Probability that the First Coin Lands Heads?

Yu — Thu, 10 Oct 2019 03:56:45 +0000

Problem 737

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.

Independent and Dependent Events of Three Coins Tossing

Yu — Fri, 28 Jun 2019 14:32:11 +0000

Problem 733

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

Recall that events $E$ and $F$ are said to be independent if
\[P(E \cap F) = P(E) P(F).\] Otherwise, they are dependent.

Solution.

First of all, we have $P(H_1)= P(H_2)= 1/2$. To calculate the probability $P(E)$, note that we have $E = \{\text{hht}, \text{thh}\}$.
Here $\text{hht}$ means that the first and the second coins land heads and the third lands tails. Similarly for $\text{thh}$.

Thus,
\[P(E)= \frac{2}{8} = \frac{1}{4}.\]

Now we consider intersections of events.
First, since $H_1 \cap H_2 = \{\text{hhh}, \text{hht}\}$, we see that
\[P(H_1 \cap H_2) = \frac{2}{8} = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} = P(H_1)\cdot P(H_2).\] Therefore, the events $H_1$ and $H_2$ are independent.

Next, as $H_1 \cap E = \{\text{hht}\}$, we have
\[P(H_1 \cap E) = \frac{1}{8} = \frac{1}{2} \cdot \frac{1}{4} = P(H_1) \cdot P(E).\] Hence, the events $H_1$ and $E$ are independent.

Finally, since $H_2 \cap E = \{\text{hht}, \text{thh}\}$, we have
\[P(H_2 \cap E) = \frac{2}{8} = \frac{1}{4}.\]

On the other hand, we have
\[P(H_2) \cdot P(E) = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}.\] It follows that $P(H_2 \cap E) \neq P(H_2) \cdot P(E)$.

Thus we conclude that the events $H_2$ and $E$ are dependent.

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