A fair coin with $1$ marked on one face and $6$ on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is $3$.
A fair coin with $1$ marked on one face and $6$ on the other and a fair die are both tossed. find the probability that the sum of numbers that turn up is $3$.
since the fair coin has $1$ marked on one face and $6$ on the other, and the die has six faces that are numbered $1,\,2,\,3\,,4,\,5,$ and $6,$ the sample space is given by
$S =\{(1,1),(1,2),(1,3),(1,4)$, $(1,5),(1,6),(6,1)$, $(6,2),(6,3),(6,4),(6,5),(6,6)\}$
Accordingly, $n ( S )=12$
Let $A$ be the event in which the sum of numbers that turn up is $3$.
Accordingly, $A=\{(1,2)\}$
$\therefore P(A)=\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}=\frac{1}{12}$
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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