Independent and Dependent Events of Three Coins Tossing

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.

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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