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# 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.

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## 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,$\begin{array}{rcl}& ⇒& \mathrm{S}=\left\{\mathrm{H},\mathrm{T}\right\}×\left\{1,2,3,4,5,6\right\}\\ \mathrm{S}& =& \left\{\begin{array}{c}\left(\mathrm{H},1\right),\left(\mathrm{H},2\right),\left(\mathrm{H},3\right),\left(\mathrm{H},4\right),\left(\mathrm{H},5\right),\left(\mathrm{H},6\right),\\ \left(\mathrm{T},1\right),\left(\mathrm{T},2\right),\left(\mathrm{T},3\right),\left(\mathrm{T},4\right),\left(\mathrm{T},5\right),\left(\mathrm{T},6\right)\end{array}\right\}\end{array}$Hence, the sample space is $\begin{array}{rcl}\mathbf{S}& \mathbf{=}& \left\{\begin{array}{c}\left(H,1\right),\left(H,2\right),\left(H,3\right),\left(H,4\right),\left(H,5\right),\left(H,6\right),\\ \left(T,1\right),\left(T,2\right),\left(T,3\right),\left(T,4\right),\left(T,5\right),\left(T,6\right)\end{array}\right\}.\end{array}$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$⇒\mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(\mathrm{E}\right)=\frac{\text{number of elemnets in}\mathrm{E}}{\text{number of elemnets in}\mathrm{S}}$The sample space $\mathrm{S}$ of this experiment is $\mathrm{S}=\left\{\begin{array}{l}\left(\mathrm{H},1\right),\left(\mathrm{H},2\right),\left(\mathrm{H},3\right),\left(\mathrm{H},4\right),\left(\mathrm{H},5\right),\left(\mathrm{H},6\right),\\ \left(\mathrm{T},1\right),\left(\mathrm{T},2\right),\left(\mathrm{T},3\right),\left(\mathrm{T},4\right),\left(\mathrm{T},5\right),\left(\mathrm{T},6\right)\end{array}\right\}$ contains twelve outcomes, and the event $\mathrm{E}$ of getting "heads and even number" contains three outcomes, $\left\{\left(\mathrm{H},2\right),\left(\mathrm{H},4\right),\left(\mathrm{H},6\right)\right\}$, so by the definition of probability,$\begin{array}{rcl}& ⇒& \mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ & =& \frac{3}{12}\\ & =& \frac{1}{4}\end{array}$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$\begin{array}{rcl}& ⇒& \mathrm{P}\left(\mathrm{F}\right)=\frac{\mathrm{n}\left(\mathrm{F}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ \mathrm{P}\left(\mathrm{F}\right)& =& \frac{\text{number of elemnets in}\mathrm{F}}{\text{number of elemnets in}\mathrm{S}}\end{array}$ The sample space of this experiment is $\mathrm{S}=\left\{\begin{array}{l}\left(\mathrm{H},1\right),\left(\mathrm{H},2\right),\left(\mathrm{H},3\right),\left(\mathrm{H},4\right),\left(\mathrm{H},5\right),\left(\mathrm{H},6\right),\\ \left(\mathrm{T},1\right),\left(\mathrm{T},2\right),\left(\mathrm{T},3\right),\left(\mathrm{T},4\right),\left(\mathrm{T},5\right),\left(\mathrm{T},6\right)\end{array}\right\}$.It contains twelve outcomes, and the event $\mathrm{F}$ of getting "heads and a number greater than $4$" contains two outcome, $\left\{\left(\mathrm{H},5\right),\left(\mathrm{H},6\right)\right\}$, so by the definition of probability,$\begin{array}{rcl}& ⇒& \mathrm{P}\left(\mathrm{F}\right)=\frac{\mathrm{n}\left(\mathrm{F}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ & =& \frac{2}{12}\\ & =& \frac{1}{6}\end{array}$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$\begin{array}{rcl}& ⇒& \mathrm{P}\left(\mathrm{G}\right)=\frac{\mathrm{n}\left(\mathrm{G}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ \mathrm{P}\left(\mathrm{G}\right)& =& \frac{\text{number of elemnets in}\mathrm{G}}{\text{number of elemnets in}\mathrm{S}}\end{array}$The sample space $\mathrm{S}$ of this experiment is $\mathrm{S}=\left\{\begin{array}{l}\left(\mathrm{H},1\right),\left(\mathrm{H},2\right),\left(\mathrm{H},3\right),\left(\mathrm{H},4\right),\left(\mathrm{H},5\right),\left(\mathrm{H},6\right),\\ \left(\mathrm{T},1\right),\left(\mathrm{T},2\right),\left(\mathrm{T},3\right),\left(\mathrm{T},4\right),\left(\mathrm{T},5\right),\left(\mathrm{T},6\right)\end{array}\right\}$ contains twelve outcomes, and the event $\mathrm{G}$ of getting "tails and odd number" contains three outcome, $\left\{\left(\mathrm{T},1\right),\left(\mathrm{T},3\right),\left(\mathrm{T},5\right)\right\}$, so by the definition of probability,$\begin{array}{rcl}& ⇒& \mathrm{P}\left(\mathrm{G}\right)=\frac{\mathrm{n}\left(\mathrm{G}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ & =& \frac{3}{12}\\ & =& \frac{1}{4}\end{array}$Hence, the probability of getting tails and an odd number is $\frac{\mathbf{1}}{\mathbf{4}}.$Hence,a) The sample space is $\begin{array}{rcl}\mathbf{S}& \mathbf{=}& \mathbf{\left\{}\begin{array}{c}\mathbf{\left(}\mathbf{H}\mathbf{,}\mathbf{1}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{H}\mathbf{,}\mathbf{2}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{H}\mathbf{,}\mathbf{3}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{H}\mathbf{,}\mathbf{4}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{H}\mathbf{,}\mathbf{5}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{H}\mathbf{,}\mathbf{6}\mathbf{\right)}\mathbf{,}\\ \mathbf{\left(}\mathbf{T}\mathbf{,}\mathbf{1}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{T}\mathbf{,}\mathbf{2}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{T}\mathbf{,}\mathbf{3}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{T}\mathbf{,}\mathbf{4}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{T}\mathbf{,}\mathbf{5}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{T}\mathbf{,}\mathbf{6}\mathbf{\right)}\end{array}\mathbf{\right\}}\end{array}$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 $\frac{\mathbf{1}}{\mathbf{4}}.$

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