Question

Out of ($2n+1$) tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is $\frac{30+k}{61}$. Find the value of $k$.

Answer 1

Let $E_1$ denote the event of $A$'s throwing ${}^{\prime }{6}^{\prime }$ and $E_2$ the event of $B$'s throwing ${}^{\prime }{7}^{\prime }$ with a pair of dice.

Then ${\overline{E}}_1,{\overline{E}}_2$ are the complementary events. There are $5$ ways of obtaining $6$ namely $(1,5),(2,4),(3,3),(4,2),(5,1)$

and similarly there are $6$ ways of getting $7$ namely $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$

$\therefore P\left(E_1\right)=\frac{5}{36}$

and $P\left(\overline{E\overline{1}}\right)=1-\frac{5}{36}=\frac{31}{36}P\left(E_2\right)=\frac{6}{36}=\frac{1}{6}$

and $P\left(\overline{E_2}\right)=1-\frac{1}{6}=\frac{5}{6}$

It is given that $A$ starts the game and he will win in the following mutually exclusive ways.

(i)$E_1$ happens i.e $A$ wins at the first draw

(ii)${\overline{E}}_1\cap \overline{E_2}\cap E_1$ happens i.e. $A$ wins at the third draw when both $A$ and $B$ fail at 1st and second draw.

(iii)${\overline{E}}_1\cap {\overline{E}}_2\cap {\overline{E}}_1\cap {\overline{E}}_2\cap {\overline{E}}_1$

happens i.e. $A$ wins at the fifth draw when both $A$ and $B$ fail at Ist, IInd, IIIrd and IVth draw and so on..........

Hence the required probability of '$A$' winning say $P\left(A\right)$ is given by $P\left(A\right)=P\left(i\right)+P\left(ii\right)+P\left(iii\right)+...$

$=P\left(E_1\right)+P({\overline{E}}_1\cap {\overline{E}}_2\cap {\overline{E}}_1)+P({\overline{E}}_1\cap {\overline{E}}_2\cap {\overline{E}}_1\cap {\overline{E}}_2\cap E)+...$

$=P\left(E_1\right)+P\left({\overline{E}}_1\right)P\left({\overline{E}}_2\right)+P\left({\overline{E}}_1\right)P\left({\overline{E}}_2\right)P\left({\overline{E}}_1\right)P{\overline{E}}_{2}P\left({\overline{E}}_1\right)$

$=\frac{5}{36}+\frac{31}{36}\cdot \frac{5}{6}\cdot \frac{5}{36}+\frac{31}{36}\cdot \frac{5}{6}\cdot \frac{31}{36}\cdot \frac{5}{6}\cdot \frac{5}{36}+....$

$=\frac{5}{36}+{(\frac{31}{36}\cdot \frac{5}{6})}^2\cdot \frac{5}{36}+...$

$=\frac{5}{36}\cdot \frac{1}{1-(\frac{31}{36}\cdot \frac{5}{6})}=\frac{5}{36}\cdot \frac{216}{61}=\frac{30}{61}$