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Suppose you are playing a game that costs $ $8 to play. You flip 10 coins and, for every head, you win $2. Whats the probability you lose money ?

$$ \begin{array}{c|cccccccccc} x& 1 & 2 & 3 & 4 & 5 & 6 & 7 &8 &9&10 \\ \hline p(x) & 5/512 & 45/1024 & 15/128 & 63/256 & 105/512 & \end{array} $$ Since the loosing money means getting less than 4 heads I just did an addition of the $$P_X(1)+P_X(2)+P_X(3)=0.1708$$


With pmf of X being $P_X(x)=\binom{10}{x}(0.5)^x(0.5)^{10-x}$


And g(X)=Y=-8+2X the equation that define the net losses or winnings.

can this be a viable solution the way I presented it ? If yes do you know any good shortcut for this exercise ?

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Don't forget about the possibility of a super bad luck event, that is you might get all tails. probability of losing money is equal to

$$\frac{1}{2^{10}}\sum_{i=0}^3 \binom{10}{i}= \frac{1+10+45+120}{1024}\approx 0.1719$$

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  • $\begingroup$ So what you did in your solution was to choose 0 or 1 or 3 and multiply it by the chances of getting heads? Moreover for my solution if I add the probability of getting 0 head I have a correct solution then. $\endgroup$ – Mahamad A. Kanouté Mar 18 '18 at 2:45
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    $\begingroup$ I did almost exactly the same thing as you isn't it? just that I included the worst case scenario. $\endgroup$ – Siong Thye Goh Mar 18 '18 at 2:48

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