Chapter 4.2, Problem 18E

(a)

To determine

The probability of nonoccurrence of event $A$.

Answer to Problem 18E

Solution: The probability of complementary event of $A$ is 0.583.

Explanation of Solution

The nonoccurrence of event $A$ will represent the complementary event of $A$ denoted as $\overline{A}$ and its probability is computed as

$P\left(A\right)+P\left(\overline{A}\right)=1$

So

$\begin{array}{c}P\left(\overline{A}\right)=1-P\left(A\right)\\ =1-0.417\\ =0.583\end{array}$

So, the probability that $A$ does not occur is 0.583.

(b)

To determine

To find: The probability that all four tosses result in the same outcome.

Answer to Problem 18E

Solution: The probability of obtaining similar outcomes is 0.125, and is obtained by the help of addition rule of probability.

Explanation of Solution

Calculation:

If a coin is tossed 4 times, then there are a total of $2^4=16$ possible outcomes in the sample space. The favorable outcomes to obtain same outcomes are $\left\{\left(HHHH\right),\left(TTTT\right)\right\}$. So, the probability that all the outcomes are same is computed by the addition theorem of probability as

$P\left\{\left(HHHH\right)\cup \left(TTTT\right)\right\}=P\left(HHHH\right)+P\left(TTTT\right)-P\left\{\left(HHHH\right)\cap \left(TTTT\right)\right\}$ $$

$P\left\{\left(HHHH\right)\cap \left(TTTT\right)\right\}=0$ because the events are independent.

So

$\begin{array}{c}P\left\{\left(HHHH\right)\cup \left(TTTT\right)\right\}=P\left(HHHH\right)+P\left(TTTT\right)\\ =\frac{1}{16}+\frac{1}{16}\\ =\frac{1}{8}\\ =0.125\end{array}$

(c)

To determine

To find: The probability of obtaining at least one head and at least one tail.

Answer to Problem 18E

Solution: The probability of at least one head and at least one tail is 0.875 and is obtained by complement rule of probability.

Explanation of Solution

Calculation:

The events favorable to getting at least 1 head and at least one tail is

$\left\{\begin{array}{l}\left(HTTT\right),\left(THTT\right),\left(TTHT\right),\left(TTTH\right),\\ \left(HHTT\right),\left(HTHT\right),\left(TTHH\right),\left(THTH\right),\\ \left(HHHT\right),\left(HHTH\right),\left(HTHH\right),\left(THHH\right)\end{array}\right\}$

The probability is obtained as

$\begin{array}{c}P\left[\text{At least one head and one tail}\right]=1-P\left[\text{0 Heads and 0 Tails}\right]\\ =1-\left[P\left(HHHH\right)+P\left(TTTT\right)\right]\\ =1-0.125\\ =0.875\end{array}$

(d)

To determine

To find: If the events $A$ and $B$ are disjoint.

Answer to Problem 18E

Solution: The provided events are not disjoint and the property of the probability theory that probability can never exceed one is used.

Explanation of Solution

Calculation:

The probability of events $A$ and $B$ are provided as

$\begin{array}{l}P\left(A\right)=0.4\\ P\left(B\right)=0.8\end{array}$

Now, by the addition theorem of probability

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

If events $A$ and $B$ are disjoint, then the probability of $A\cap B$ must have been zero.

So

$\begin{array}{c}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)\\ =0.4+0.8\\ =1.2\end{array}$

Since the probability of any event can never be more than 1, hence $A$ and $B$ are not disjoint.

(e)

To determine

If the probability of an event A can be −0.04.

Answer to Problem 18E

Solution: The provided scenario is not possible and the property of the probability theory that the probability can never be less than 0 is used.

Explanation of Solution

According to the provided information, the event $A$ is very rare and its probability is $-0.04$. This is not possible because the probability of any event can never be negative. Probability always lies in between 0 and 1.