A die is thrown, find the probability of following events: A number more than $6$ will appear,
A die is thrown, find the probability of following events: A number more than $6$ will appear,
The sample space of the given experiment is given by
$S=\{1,2,3,4,5,6\}$
Let $D$ be the event of the occurrence of a number greater than $6.$
Accordingly, $D=\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{I}$
$\therefore P(D)=\frac{\text { Number of outcomes favourable to } D}{\text { Total number of possible outcomes }}=\frac{n(D)}{n(S)}=\frac{0}{6}=0$
Similar Questions
Let $\mathrm{X}$ and $\mathrm{Y}$ be two events such that $\mathrm{P}(\mathrm{X})=\frac{1}{3}, \mathrm{P}(\mathrm{X} \mid \mathrm{Y})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{Y} \mid \mathrm{X})=\frac{2}{5}$. Then
$[A]$ $\mathrm{P}\left(\mathrm{X}^{\prime} \mid \mathrm{Y}\right)=\frac{1}{2}$ $[B]$ $\mathrm{P}(\mathrm{X} \cap \mathrm{Y})=\frac{1}{5}$ $[C]$ $\mathrm{P}(\mathrm{X} \cup \mathrm{Y})=\frac{2}{5}$ $[D]$ $\mathrm{P}(\mathrm{Y})=\frac{4}{15}$
Let $\mathrm{X}$ and $\mathrm{Y}$ be two events such that $\mathrm{P}(\mathrm{X})=\frac{1}{3}, \mathrm{P}(\mathrm{X} \mid \mathrm{Y})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{Y} \mid \mathrm{X})=\frac{2}{5}$. Then
$[A]$ $\mathrm{P}\left(\mathrm{X}^{\prime} \mid \mathrm{Y}\right)=\frac{1}{2}$ $[B]$ $\mathrm{P}(\mathrm{X} \cap \mathrm{Y})=\frac{1}{5}$ $[C]$ $\mathrm{P}(\mathrm{X} \cup \mathrm{Y})=\frac{2}{5}$ $[D]$ $\mathrm{P}(\mathrm{Y})=\frac{4}{15}$
- [IIT 2017]
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