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The question asks:

Suppose that in a weekly lottery you have probability .02 of winning a prize with a single ticket. If you buy 1 ticket per week for 52 weeks, what is the probability that you win 3 or more prizes?

I realized that the probability function of this problem would can be represented by a binomial distribution such that if we let $x =$ number of prizes won then the probability that I win $3$ or more prizes can be represented by the cumulative distribution function $$P(X\ge3) = \binom{52}{x}(0.02^x)(0.98^{52-x})$$

To find $x$ I realized that to satisfy the condition that $X\ge x$ I needed $x \in \{3,4,5...,52\}$ since there are $50$ possible numbers that $x$ can be I let $x = 50$. The answer should be $0.0859$ but I cannot seem to get that answer.

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4 Answers 4

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First, notice that you mis-wrote the Binomial distribution. For some $x$, you should have ${52\choose x}(0.02^{x})(0.98^{52-x})$, and not the power of $x$ for both your successes and failures (evidently).

Then much easier to calculate is realizing that this is equivalent to $$ 1 - P(\text{win less than 3 prizes})$$ which is given by $$1 - \sum_{x=0}^{2}{52\choose x}(0.02^{x})(0.98^{52-x})$$

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  • $\begingroup$ You're completely right I'm not sure what I was thinking I have changed the formula. Is there another way of solving this problem besides using the complement of $P(\text {win less than 3 prizes})$? $\endgroup$
    – Daniel
    Aug 24, 2015 at 15:19
  • $\begingroup$ @Daniel Well, you could calculate directly without using the complement of course. But that is a lot more computation! $\endgroup$
    – miradulo
    Aug 24, 2015 at 15:20
  • $\begingroup$ By calculating directly what would the value of $x$ be? I seem to be confused in finding the correct $x$ value. Would it mean I would have to calculate $x$ for each different value $\ge3$? $\endgroup$
    – Daniel
    Aug 24, 2015 at 15:23
  • $\begingroup$ @Daniel That's what I mean, yes. A similar summation, but for $x \geq 3$ to your value of $52$. If you are solving by hand, this question was almost certainly meant to be calculated using the complement. $\endgroup$
    – miradulo
    Aug 24, 2015 at 15:25
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The correct formula for $P(X)$ is $$ P(X = x) = \binom{52}{x}(0.02)^x(0.98)^{52-x}. $$ Then the formula for $P(X \ge 3)$ is, $$ P(X \ge 3) = \sum_{x = 3}^{52} \binom{52}{x}(0.02)^x(0.98)^{52-x}. $$ Of course, if you don't have a computer available, it's going to be much more feasible to compute $1 - P(X \ge 3) = P(X \le 2)$, as Donkey Kong suggests.

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You have to use the cummulative distribution.

$P(X \geq 3)=\sum_{k=3}^{52} {52 \choose k} \cdot 0.02^k\cdot0.98^{52-k}$

If you calculate like this then you have to sum up 50 terms. A better way is to use the converse probability.

$1-P(X \leq 2)=1-\sum_{k=0}^{2} {52 \choose k} \cdot 0.02^k\cdot 0.98^{52-k}\approx 0.0859$

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As you said, this can be modeled by binomial distribution.

The answer should be

$$ P(X \ge 3) = 1 - (P( X = 0) + P(X = 1) + P(X = 2)) $$

$$ P(X = 0) = { 52 \choose 0 } 0.02^0 * 0.98^{52} = 0.34975 $$

$$ P(X = 1) = { 52 \choose 1 } 0.02^1 * 0.98^{51} = 0.371162 $$

$$ P(X = 2) = { 52 \choose 2 } 0.02^2 * 0.98^{50} = 0.19315 $$

Therefore, $$ P(X \ge 3) = 1 - (0.34975 + 0.371162 + 0.19315) = 0.085938 $$

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