Question

An experiment consists of tossing a coin and rolling a die.
(a) Find the sample space.
(b) Find the probability of getting heads and an even number.
(c) Find the probability of getting heads and a number greater than.$4.$
(d) Find the probability of getting tails and an odd number.

Answer 1

Step- 1: Find the sample space:

Given an experiment consists of tossing a coin and rolling a die.

If an experiment consists of tossing a coin and rolling a die and recording the results in order, the sample space of an experiment is the set of all possible outcomes.

If we let $\mathrm{H}$ stands for heads and $\mathrm{T}$ for tails, then the sample space of the tossing a coin and rolling a dying experiment is,

$\begin{array}{rcl}& \Rightarrow & \mathrm{S}=\{\mathrm{H},\mathrm{T}\}\times \{1,2,3,4,5,6\}\\ \mathrm{S}& =& \left\{\begin{array}{c}(\mathrm{H},1),(\mathrm{H},2),(\mathrm{H},3),(\mathrm{H},4),(\mathrm{H},5),(\mathrm{H},6),\\ (\mathrm{T},1),(\mathrm{T},2),(\mathrm{T},3),(\mathrm{T},4),(\mathrm{T},5),(\mathrm{T},6)\end{array}\right\}\end{array}$

Hence, the sample space is $\begin{array}{rcl}\mathbf{S}& \mathbf{=}& \left\{\begin{array}{c}(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),\\ (T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\end{array}\right\}.\end{array}$

Step-2 : Find the probability of getting heads and an even number:

Given an experiment consists of tossing a coin and rolling a die.

Let $\mathrm{S}$ be the sample space of an experiment in which all outcomes are equally likely and let $\mathrm{E}$ be an event.

The probability of $\mathrm{E}$, written $\mathrm{P}\left(\mathrm{E}\right)$, is
$$\begin{gathered} \Rightarrow \mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ \mathrm{P}\left(\mathrm{E}\right)=\frac{\text{number of elemnets in}\mathrm{E}}{\text{number of elemnets in}\mathrm{S}} \end{gathered}$$

The sample space $\mathrm{S}$ of this experiment is $\mathrm{S}=\left\{\begin{array}{l}(\mathrm{H},1),(\mathrm{H},2),(\mathrm{H},3),(\mathrm{H},4),(\mathrm{H},5),(\mathrm{H},6),\\ (\mathrm{T},1),(\mathrm{T},2),(\mathrm{T},3),(\mathrm{T},4),(\mathrm{T},5),(\mathrm{T},6)\end{array}\right\}$ contains twelve outcomes, and the event $\mathrm{E}$ of getting "heads and even number" contains three outcomes, $\left\{\right(\mathrm{H},2),(\mathrm{H},4),(\mathrm{H},6\left)\right\}$, so by the definition of probability,
$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ & =& \frac{3}{12}\\ & =& \frac{1}{4}\end{array}$
Hence, the probability of getting heads and an even number is $\frac{\mathbf{1}}{\mathbf{4}}.$

Step-3: Find the probability of getting heads and a number greater than $\mathbf{4}\mathbf{:}$

Let $\mathrm{S}$ be the sample space of an experiment in which all outcomes are equally likely and let $\mathrm{F}$ be an event. The probability of $\mathrm{F}$, written $\mathrm{P}\left(\mathrm{F}\right)$, is
$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{F}\right)=\frac{\mathrm{n}\left(\mathrm{F}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ \mathrm{P}\left(\mathrm{F}\right)& =& \frac{\text{number of elemnets in}\mathrm{F}}{\text{number of elemnets in}\mathrm{S}}\end{array}$
The sample space of this experiment is $\mathrm{S}=\left\{\begin{array}{l}(\mathrm{H},1),(\mathrm{H},2),(\mathrm{H},3),(\mathrm{H},4),(\mathrm{H},5),(\mathrm{H},6),\\ (\mathrm{T},1),(\mathrm{T},2),(\mathrm{T},3),(\mathrm{T},4),(\mathrm{T},5),(\mathrm{T},6)\end{array}\right\}$.
It contains twelve outcomes, and the event $\mathrm{F}$ of getting "heads and a number greater than $4$" contains two outcome, $\left\{\right(\mathrm{H},5),(\mathrm{H},6\left)\right\}$, so by the definition of probability,
$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{F}\right)=\frac{\mathrm{n}\left(\mathrm{F}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ & =& \frac{2}{12}\\ & =& \frac{1}{6}\end{array}$
Hence, the probability of getting heads and a number greater than $\mathbf{4}$ is $\frac{\mathbf{1}}{\mathbf{6}}.$

Step- 4: Find the probability of getting tails and an odd number:

Let $\mathrm{S}$ be the sample space of an experiment in which all outcomes are equally likely and let $\mathrm{G}$ be an event.

The probability of $\mathrm{G}$, written $\mathrm{P}\left(\mathrm{G}\right)$, is
$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{G}\right)=\frac{\mathrm{n}\left(\mathrm{G}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ \mathrm{P}\left(\mathrm{G}\right)& =& \frac{\text{number of elemnets in}\mathrm{G}}{\text{number of elemnets in}\mathrm{S}}\end{array}$

The sample space $\mathrm{S}$ of this experiment is $\mathrm{S}=\left\{\begin{array}{l}(\mathrm{H},1),(\mathrm{H},2),(\mathrm{H},3),(\mathrm{H},4),(\mathrm{H},5),(\mathrm{H},6),\\ (\mathrm{T},1),(\mathrm{T},2),(\mathrm{T},3),(\mathrm{T},4),(\mathrm{T},5),(\mathrm{T},6)\end{array}\right\}$ contains twelve outcomes, and the event $\mathrm{G}$ of getting "tails and odd number" contains three outcome, $\left\{\right(\mathrm{T},1),(\mathrm{T},3),(\mathrm{T},5\left)\right\}$, so by the definition of probability,
$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(\mathrm{G}\right)=\frac{\mathrm{n}\left(\mathrm{G}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ & =& \frac{3}{12}\\ & =& \frac{1}{4}\end{array}$
Hence, the probability of getting tails and an odd number is $\frac{\mathbf{1}}{\mathbf{4}}.$

Hence,

a) The sample space is $\begin{array}{rcl}\mathbf{S}& \mathbf{=}& \mathbf{\left\{}\begin{array}{c}\mathbf{(}\mathbf{H}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{H}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{H}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{H}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{H}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{H}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{,}\\ \mathbf{(}\mathbf{T}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{T}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{T}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{T}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{T}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{T}\mathbf{,}\mathbf{6}\mathbf{)}\end{array}\mathbf{\right\}}\end{array}$

b) The probability of getting heads and an even number is $\frac{\mathbf{1}}{\mathbf{4}}.$

c) The probability of getting heads and a number greater than $\mathbf{4}$ is $\frac{\mathbf{1}}{\mathbf{6}}.$

d) The probability of getting tails and an odd number is $\frac{\mathbf{1}}{\mathbf{4}}.$