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It is believed that $71\%$ of tickets purchased for an upcoming music festival are actually legitimate tickets. The rest are fake tickets sold by scam artists. Assume everyone who bought a ticket shows up at the concert. $\newline$ If security guards at the festival randomly select $2$ tickets to examine in more detail, what is the probability that exactly $2$ of the tickets are legitimate? $\newline$ Write your answer as a decimal rounded to the nearest thousandth. $\newline$ ____ $\newline$

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Find probabilities using the binomial distribution

Full solution

Q. It is believed that

71\%

of tickets purchased for an upcoming music festival are actually legitimate tickets. The rest are fake tickets sold by scam artists. Assume everyone who bought a ticket shows up at the concert.

\newline

If security guards at the festival randomly select

2

tickets to examine in more detail, what is the probability that exactly

2

of the tickets are legitimate?

\newline

Write your answer as a decimal rounded to the nearest thousandth.

\newline

____

\newline

Calculate First Ticket Probability: question_prompt: What is the probability that exactly $2$ of the randomly selected tickets are legitimate?
Calculate Second Ticket Probability: Step $1$ : Calculate the probability of the first ticket being legitimate. Since $71\%$ of the tickets are legitimate, the probability is $0.71$ .
Calculate Probability of Both Tickets: Step $2$ : Calculate the probability of the second ticket being legitimate. This is also $0.71$ , because the probability doesn't change for the second ticket.
Round to Nearest Thousandth: Step $3$ : Multiply the probabilities from step $1$ and step $2$ to find the probability that both tickets are legitimate. So, $0.71 \times 0.71 = 0.5041$ .
Round to Nearest Thousandth: Step $3$ : Multiply the probabilities from step $1$ and step $2$ to find the probability that both tickets are legitimate. So, $0.71 \times 0.71 = 0.5041$ . Step $4$ : Round the result to the nearest thousandth as instructed. The rounded probability is $0.504$ .

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