Chapter 4.2, Problem 20E

(a)

To determine

To find: The probability of nonoccurrence of event $A$.

Answer to Problem 20E

Solution: The probability of complementary event of $A$ is 0.776.

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

$\begin{array}{c}P\left(\overline{A}\right)=1-P\left(A\right)\\ =1-0.224\\ =0.776\end{array}$

So, the probability that $A$ does not occur is 0.776.

(b)

To determine

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

Answer to Problem 20E

Solution: The probability of obtaining similar outcomes is 0.25, and is obtained by the help of addition rule of probability.

Explanation of Solution

Calculation:

If a coin is tossed threetimes, then there are a total of $2^3=8$ possible outcomes in the sample space. The favorable outcomes to obtain same outcomes are $\left\{\left(HHH\right),\left(TTT\right)\right\}$. So, the probability that all the outcomes are same is computed by the addition theorem of probability as

$P\left\{\left(HHH\right)\cup \left(TTT\right)\right\}=P\left(HHH\right)+P\left(TTT\right)-P\left\{\left(HHH\right)\cap \left(TTT\right)\right\}$$$

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

So

$\begin{array}{c}P\left\{\left(HHH\right)\cup \left(TTT\right)\right\}=P\left(HHH\right)+P\left(TTT\right)\\ =\frac{1}{8}+\frac{1}{8}\\ =\frac{1}{4}\\ =0.25\end{array}$

(c)

To determine

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

Answer to Problem 20E

Solution: The probability of at least one head and at least one tail is 0.75 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

$\left\{\begin{array}{l}\left(HHT\right),\left(HTH\right),\left(TTH\right),\\ \left(HTT\right),\left(THH\right),\left(THT\right)\end{array}\right\}$

The probability is obtained as

$\begin{array}{c}P\left[\text{At least one head and one tail}\right]=1-P\left[\text{0 Heads and 0 Tails}\right]\\ =1-\left[P\left(HHH\right)+P\left(TTT\right)\right]\\ =1-0.25\\ =0.75\end{array}$

(d)

To determine

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

Answer to Problem 20E

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

$\begin{array}{l}P\left(A\right)=0.5\\ P\left(B\right)=0.6\end{array}$

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

$\begin{array}{c}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)\\ =0.5+0.6\\ =1.1\end{array}$

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

(e)

To determine

To explain: If the probability of an event A can be −0.01.

Answer to Problem 20E

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 $-0.01$. This is not possible because the probability of any event can never be negative. Probability always lies in between 0 and 1.