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In a game, a man wins $Rs.\,100$ if he gets $5$ or $6$ on a throw of a fair die and loses $Rs.\,50$ for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is
$\frac{{400}}{9}\,loss$
$0$
$\frac{{400}}{3}\,gain$
$\frac{{400}}{3}\,loss$
Solution
Let $w$ denotes probability that outcome $5$ or $6\left(w=\frac{2}{6}=\frac{1}{3}\right)$
Let, $L$ denotes probability that outcome $1,2,3,4\left(L=\frac{4}{6}=\frac{2}{3}\right)$
Expected Gain/Loss
$=\mathrm{w} \times 100+\mathrm{Lw}(-50+100)+\mathrm{L}^{2} \mathrm{w}(-50-50+100)+\mathrm{L}^{3}(-150)$
$=\frac{1}{3} \times 100+\frac{2}{3} \cdot \frac{1}{3}(50)+\left(\frac{2}{3}\right)^{2}\left(\frac{1}{3}\right)(0)+\left(\frac{2}{3}\right)^{3}(-150)=0$
