In a lottery, the top cash prize was $\$678$ million, going to three lucky winners. Players pick five different numbers from $1$ to $56$ and one number from $1$ to $44$ A player wins a minimum award of $\$450$ by correctly matching three numbers drawn from the white balls ( $1$ through $56$ ) and matching the number on the gold ball ( $1$ through $44$ ). What is the probability of winning the minimum award? The probability of winning the minimum award is $1$ $0$ (Type an integer or a simplified fraction.)

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Precalculus

Find probabilities using the binomial distribution

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Q. In a lottery, the top cash prize was

\$678

million, going to three lucky winners. Players pick five different numbers from

1

56

and one number from

1

44

A player wins a minimum award of

\$450

by correctly matching three numbers drawn from the white balls (

1

through

56

) and matching the number on the gold ball (

1

through

44

). What is the probability of winning the minimum award? The probability of winning the minimum award is

1

0

(Type an integer or a simplified fraction.)

Calculate Probability: To calculate the probability of winning the minimum award, we need to determine the number of ways to correctly match three numbers from the white balls and one number from the gold ball. $\newline$ First, we calculate the number of ways to choose three correct white balls from the five that are drawn. $\newline$ There are $56$ white balls, and players pick $5$ of them, so there are $C(56, 5)$ ways to pick any $5$ white balls, where $C(n, k)$ is the combination of $n$ items taken $k$ at a time.
Choose Correct White Balls: Next, we calculate the number of ways to choose the three correct white balls from the five that are drawn. Since the order in which the three correct balls are drawn doesn't matter, we use combinations again. There are $C(5, 3)$ ways to choose three correct white balls from the five that are drawn.
Choose Incorrect White Balls: Now, we calculate the number of ways to choose the two incorrect white balls from the remaining $51$ white balls ( $56$ total white balls minus the $5$ correct ones). $\newline$ There are $C(51, 2)$ ways to choose these two incorrect white balls.
Choose Correct Gold Ball: For the gold ball, there is only one correct number out of $44$ possible numbers. $\newline$ So, there is only $1$ way to choose the correct gold ball.
Total Possible Outcomes: The total number of possible outcomes for the white balls is $C(56, 5)$ , as calculated in the first step.
Calculate Probability Formula: The total number of possible outcomes for the gold ball is $44$ , since there are $44$ possible numbers to choose from.
Perform Calculations: Now, we calculate the probability of winning the minimum award by multiplying the number of ways to get the correct combination of white and gold balls and then dividing by the total number of possible outcomes. $\newline$ The probability is $\frac{C(5, 3) \times C(51, 2) \times 1}{C(56, 5) \times 44}$ .
Plug Values: We perform the calculations using the combination formula $C(n, k) = \frac{n!}{k! * (n - k)!}$ , where $n!$ is the factorial of $n$ .
$C(5, 3) = \frac{5!}{3! * (5 - 3)!} = \frac{(5 * 4)}{(2 * 1)} = 10$
$C(51, 2) = \frac{51!}{2! * (51 - 2)!} = \frac{(51 * 50)}{(2 * 1)} = 1275$
$C(56, 5) = \frac{56!}{5! * (56 - 5)!} = \frac{(56 * 55 * 54 * 53 * 52)}{(5 * 4 * 3 * 2 * 1)} = 2598960$
Simplify Fraction: Now we plug these values into the probability formula: $\newline$ Probability = $(10 \times 1275 \times 1) / (2598960 \times 44) = 12750 / 114354240$
Simplify Fraction: Now we plug these values into the probability formula: $\newline$ Probability = $(10 \times 1275 \times 1) / (2598960 \times 44) = 12750 / 114354240$ We simplify the fraction $12750 / 114354240$ to its lowest terms. $\newline$ $12750 / 114354240$ can be simplified by dividing both the numerator and the denominator by $12750$ . $\newline$ $12750 / 12750 = 1$ $\newline$ $114354240 / 12750 = 8968$ $\newline$ So, the simplified probability is $1 / 8968$ .