In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair dice is thrown. The man decided to throw a dice thrice but he quits as and when he gets a six. Find the expected value of the amount he wind loses.

: Probability of getting a six $= \cfrac{1}{6}$

$\therefore$ $q = 1 - p = 1 - \cfrac{1}{6} = \cfrac{5}{6}$

(i) Let $E$ : the man gets a six in first throw, then $P(E) = \cfrac{1}{6}$

(ii) Let $F$ : the man does not get a six in first throw,

but he gets a six in the second throw,

then $P(F) = \cfrac{5}{6} \times \cfrac{1}{6} = \cfrac{5}{{36}}$

(iii) Let $G$ : the man does not get a six in first two throws but he gets a six in the third throw,

then $P(G) = \cfrac{5}{6} \times \cfrac{5}{6} \times \cfrac{1}{6} = \cfrac{{25}}{{216}}$

(iv) Let $H$ : the man does not get a six in any ofthree throws, then

$P(H) = \cfrac{5}{6} \times \cfrac{5}{6} \times \cfrac{5}{6} = \cfrac{{125}}{{216}}$

When the man gets a six in the first throw, he gets Re.1.

When the man gets a six in the second throw,

he gets Rs. $( - 1 + 1) = {\mathop{\rm Rs}\nolimits} .0$

When the man gets a six in the third throw, he gets Rs. $( - 1 - 1 + 1) = {\rm{Rs}}.( - 1)$ i.e., he loses Re.1.

When the man gets no six in any of the three thorws he gets Rs. $( - 1 - 1 - 1) = {\rm{Rs}}.( - 3),$ i.e., he loses Rs. 3

Expected value $E(X) = \left( {\cfrac{1}{6}} \right)(1) + \left( {\cfrac{5}{{36}}} \right)(0) + \left( {\cfrac{{25}}{{216}}} \right)( - 1) + \cfrac{{125}}{{216}}( - 3)$

$= \cfrac{1}{6} - \cfrac{{25}}{{216}} - \cfrac{{375}}{{216}} = \cfrac{{36 - 25 - 375}}{{216}} = {\rm{Rs}} \cdot \cfrac{{91}}{{54}}({\rm{loss}})$