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The probability of winning a game is
1//1000. If you play this game 500 times, what is the probability that you win at least twice?

The probability of winning a game is $1 / 1000$ . If you play this game $500$ times, what is the probability that you win at least twice?

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Probability of independent and dependent events

Full solution

Q. The probability of winning a game is

1 / 1000

. If you play this game

500

times, what is the probability that you win at least twice?

Identify probability of not winning: Identify the probability of not winning the game. $\newline$ The probability of not winning the game is the complement of the probability of winning the game. $\newline$ Probability of not winning = $1 - \text{Probability of winning} = 1 - \frac{1}{1000} = \frac{999}{1000}$ .
Calculate probability of winning once: Calculate the probability of winning exactly once in $500$ tries. $\newline$ This can be done using the binomial probability formula: $\newline$ $P(X = k) = C(n, k) \cdot (p^k) \cdot ((1-p)^{(n-k)})$ $\newline$ Where: $\newline$ $P(X = k)$ is the probability of $k$ successes in $n$ trials, $\newline$ $C(n, k)$ is the number of combinations of $n$ things taken $k$ at a time, $\newline$ $p$ is the probability of success on a single trial, and $\newline$ $n$ is the number of trials. $\newline$ For winning exactly once: $\newline$ $P(X = k) = C(n, k) \cdot (p^k) \cdot ((1-p)^{(n-k)})$ $0$ $\newline$ $P(X = k) = C(n, k) \cdot (p^k) \cdot ((1-p)^{(n-k)})$ $1$
Calculate probability of never winning: Calculate the probability of never winning in $500$ tries. $\newline$ This is simply the probability of not winning raised to the power of the number of tries: $\newline$ $P(X = 0) = (\frac{999}{1000})^{500}$
Calculate probability of winning at least twice: Calculate the probability of winning at least twice. The probability of winning at least twice is the complement of the probability of winning fewer than two times (which includes winning exactly once and never winning). P(X \geq 2) = 1 - P(X < 2) P(X < 2) = P(X = 0) + P(X = 1) $P(X \geq 2) = 1 - (P(X = 0) + P(X = 1))$
Perform calculations: Perform the calculations using the results from Steps $2$ and $3$ . $\newline$ $P(X \geq 2) = 1 - ((999/1000)^{500} + 500 \times (1/1000) \times (999/1000)^{499})$
Simplify expression: Simplify the expression. This step involves arithmetic calculations to simplify the expression obtained in Step $5$ .

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