Chapter 6.3, Problem 22P

Critical Thinking Let r be a binomial random variable representing the number of successes out of n trials.

(a) Explain why the sample space for r consists of the set {0, 1, 2,…, n}

and why the sum of the probabilities of all the entries in the entire sample space must be 1.

$\begin{array}{l}\left(\text{b}\right)\text{ Explain why }P(r\ge 1)=1-P\left(0\right).\hfill \\ \left(\text{c}\right)\text{ Explain why }P(r\ge 2)=1-P(0)-P\left(1\right).\hfill \\ \left(\text{d}\right)\text{ Explain why }P(r\ge m)=1-P\left(0\right)-P(1)-\mathrm{...}-P(m-1)\text{ for }1\le m\le n.\hfill \end{array}$

To determine

To explain: The reason that the sample space consists of the set {0, 1, 2, …, n}, and the reason that the sum of the probabilities in the sample space is equal to 1.

Answer to Problem 22P

Solution: A variable, r, which represents the number of successes in n trials of an experiment. There are n trials, and so, the number of trials should be $0,1,2\dots ,n$, which represents that the possible values of r can be 0 through n.

Explanation of Solution

Given: Consider r to be a binomial random variable used to represent the number of successes out of n trials.

Calculation: There are n trials; so, the number of trials should be $0,1,2\dots ,n$, which represents that the possible values of r can be 0 through n.

The sample space is a collection of all the possible outcomes in an experiment. It is known that the sum of the probabilities of all outcomes should be equal to 1. Consider an example of tossing a coin. The sample space of tossing a coin is, $S=\left\{H,T\right\}$.

The probability of obtaining heads can be calculated as:

$\begin{array}{c}P\left(\text{obtaining heads}\right)=\frac{\text{Number of favourable outcomes}}{\text{Total outcomes}}\\ =\frac{1}{2}\end{array}$

The probability of obtaining tails can be calculated as:

$\begin{array}{c}P\left(\text{obtaining tails}\right)=\frac{\text{Number of favourable outcomes}}{\text{Total outcomes}}\\ =\frac{1}{2}\end{array}$

The sum of the probabilities of obtaining heads and tails can be calculated as:

$\begin{array}{c}P\left(\text{Sample space}\right)=\text{Probability of obtaining heads + Probability of obtaining tails}\\ \text{=}\frac{1}{2}+\frac{1}{2}\\ =1\end{array}$

Hence, the sum of the probabilities of the sample space is equal to 1.

To determine

To explain: The reason behind the probability, $P\left(r\ge 1\right)$, being equal to $1-P\left(r=0\right)$.

Answer to Problem 22P

Solution: The probability, $P\left(r\ge 1\right)$, is equal to $1-P\left(r=0\right)$ because $r\ge 1$ contains all the possible outcomes, excluding $r=0$.

Explanation of Solution

Given: Consider r to be a binomial random variable that represents the number of all successes out of n trials.

Calculation: The probability of r greater than or equal to 1 can be written as:

$P\left(r\ge 1\right)=1-P\left(r<1\right)$

Or,

$P\left(r\ge 1\right)=P\left(r=1\right)+P\left(r=2\right)+P\left(r=3\right)+\mathrm{...}+P\left(r=n\right)$

From the above, it is clear that the probability of $r\ge 1$ contains all the possible outcomes, excluding $r=0$. Thus, it can be written as:

$P\left(r\ge 1\right)=1-P\left(r=0\right)$.

To determine

To explain: The reason behind the probability, $P\left(r\ge 2\right)$, being equal to $1-P\left(r=0\right)-P\left(r=1\right)$.

Answer to Problem 22P

Solution: The probability, $P\left(r\ge 2\right)$, is equal to $1-P\left(r=0\right)-P\left(r=1\right)$ because $r\ge 2$ contains all the possible outcomes, excluding $r=0$ and r = 1.

Explanation of Solution

Given: Consider r to be a binomial random variable that represents the number of all successes out of n trials.

Calculation: The probability of r greater than or equal to 2 can be written as:

$\begin{array}{c}P\left(r\ge 2\right)=1-P\left(r<2\right)\\ =1-\left[P\left(r=0\right)+P\left(r=1\right)\right]\end{array}$

Or,

$P\left(r\ge 2\right)=P\left(r=2\right)+P\left(r=3\right)+\mathrm{...}+P\left(r=n\right)$

From the above, it is clear that the probability of $r\ge 2$ contains all the possible outcomes, excluding $r=1\text{ and }r=2$. Thus:

$P\left(r\ge 2\right)=1-P\left(r=0\right)-P\left(r=2\right)$

To determine

To explain: The reason behind the probability, $P\left(r\ge m\right)$, being equal to $1-P\left(r=0\right)-P\left(r=1\right)-\mathrm{...}-P\left(m-1\right)$ for $1\le m\le n$.

Answer to Problem 22P

Solution: The probability, $P\left(r\ge m\right)=1-\left[P\left(r=0\right)+P\left(r=1\right)+\mathrm{...}+P\left(r=m-1\right)\right]$, because $r\ge m$ contains all the possible outcomes, excluding $r=0$ to $r=m-1$.

Explanation of Solution

Given Consider r to be a binomial random variable that represents the number of all successes out of n trials.

Calculation: The probability of r greater than or equal to m can be written as:

$\begin{array}{c}P\left(r\ge m\right)=1-P\left(r<m\right)\\ =1-\left[P\left(r=0\right)+P\left(r=1\right)+\mathrm{...}+P\left(r=m-1\right)\right]\end{array}$

Or,

$P\left(r\ge m\right)=P\left(r=m\right)+P\left(r=m+1\right)$

From the above, it is clear that the probability of $r\ge m$ contains all the possible outcomes, excluding $r=0,r=1,\mathrm{....},r=m-1$. Thus, $P\left(r\ge 2\right)=1-P\left(r=0\right)-P\left(r=1\right)\mathrm{....}-P\left(r=m-1\right)$.