Four fair coins are tossed.

(1) What is the probability that all coins land heads?

(2) What is the probability that all coins land heads if the first coin is heads?

(3) What is the probability that all coins land heads if at least one coin lands heads?

Solution.

Solution (1)

There are $2^4=16$ total possible outcomes of which only one outcome gives rise to all heads.

Thus the probability that all coins land heads is $1/16$.

Solution (2)

Consider the event that the first coin is heads.

In this case, there are total $2^3=8$ possible outcomes for the rest of coins (2nd, 3rd, and 4th).

Hence, the probability that all coins land heads given that the first coin is heads is $1/8$.

Solution (3)

Let $H$ be the event that all coins land heads. Let $F$ be the event that at least one coin lands heads. Then the required conditional probability is given by
\begin{align*}
P(H \mid F) &= \frac{P(H \cap F)}{P(F)}.
\end{align*}
The complement $F^c$ of $F$ is the event that all lands tails whose probability $P(F^c)$ is $1/16$ just like part (a). Hence
\[P(F) = 1 – P(F^c) = 1 – \frac{1}{16} = \frac{15}{16}.\]

It follows that
\begin{align*}
P(H \mid F) &= \frac{P(H \cap F)}{P(F)}\\[6pt] &= \frac{P(H)}{P(F)}\\[6pt] &= \frac{1/16}{15/16}\\[6pt] &= \frac{1}{15}.
\end{align*}

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Two fair and distinguishable six-sided dice are rolled.

(1) What is the probability that the sum of the upturned faces will equal $5$?

(2) What is the probability that the outcome of the second die is strictly greater than the first die?

Solution.

The sample space $S$ is the set of all pairs $(i, j)$ with $1 \leq i, j \leq 6$, where $i$ and $j$ are the numbers corresponding to the first and the second die respectively. Hence $|S| = 36$. All of the $36$ possible outcomes are equally likely.

Solution (1)

Now we consider the outcomes that give rise to the event that the sum of the upturned faces is $5$. There are four such outcomes:
\[(1, 4), (2, 3), (3, 2), (4, 1).\] Thus, the desired probability is $\frac{4}{36}=\frac{1}{9}$.

Solution (2)

The outcomes that give rise to this event are $(i, j)$ with $i < j$, where $i$ and $j$ are the numbers corresponding to the first and second die respectively. We count the number of such outcomes as follows. When $i=1$, possible values for $j$ are $j=2, 3, 4, 5, 6$. When $i=2$, possible values for $j$ are $j = 3, 4, 5, 6$. In general, $j$ can be $j = i+1, i+2, \dots, 6$. Note that $i$ cannot be $6$, otherwise $j$ is not strictly greater than $i$. Counting this way, we see that there are \[5+4+3+2+1 = 15\] outcomes, each of probability $1/36$. Hence the probability that the outcomes of the second die is strictly greater than the first die is \[15 \times \frac{1}{36} = \frac{15}{36} = \frac{5}{12}.\]

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