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The probability of winning a lotter is $0.01$. How many times do you need to play to win with probability at least $0.6$? Assume that lottery draws are statistically independent.

My failed attempt:

Let $P(W)=0.01\\P(L)=0.99$

and let $X$ be the number of wins.

Then the event $\{X=k\}$ follows a binomial distribution. So we calculate how many trials are needed for the probability of the former event to be equal to at least $0.6$:

$P(X=k) = \binom{n}{k}(0.01)^k(0.99)^{n-k}=0.6$

I should now solve for $n$.

Is this the correct approach?

Slim Shady
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2 Answers2

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HINT: The probability to win at least once is

$$P(X\geq 1)=1-P(X=0),$$ where $X\sim Bin(n,0.01)$

Remark:

After some steps you get $0.99^n\color{blue}{\leq} 0.4$.

\begin{array} \\ & \textrm{taking logs}\\ & \\ & \ln\left(0.99^n\right) \leq \ln(0.4) \\ & \\ & n\cdot \ln\left(0.99\right) \leq \ln(0.4) \\ \end{array}

Now you divide the equation by $\ln(0.99)$. Here you have to regard that $\ln(x)<0$ if $0<x<1$. That means that you divide the equation by a factor which is negative. Consequently the inequality sign turns around.

$$n\geq \frac{\ln(0.4)}{\ln(0.99)}$$

callculus42
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$$0.01+0.99^1\times 0.01+(0.99)^2\times0.01+\cdots+0.99^{n-1}\times0.01$$

$$=0.01\times(1+0.99^1+0.99^2+\cdots+0.99^{n-1})$$

$$=0.01\times\frac{(1-0.99^n)}{1-0.99}$$

$$=1-0.99^n\geq 0.6 \Leftrightarrow 0.99^n\geq0.4$$

$$\Leftrightarrow n\log 0.99\geq \log 0.4 \Leftrightarrow n\geq \frac{\log 0.4}{\log 0.99}\Rightarrow n\geq 91.17\Rightarrow n\geq 92$$

F. A. Mala
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