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$Probability with the Question 10, 11.5.9 points Fundamental Countin Points: 0 of 1 Save Question list Question 6 Question 7 x//3 Question 8 Question 9 In a lottery, the top cash prize was $610 million, going to three lucky winners. Players pick four different numbers from 1 to 56 and one number from 1 to 43 A player wins a minimum award of $300 by correctly matching two numbers drawn from the white balls ( 1 through 56) and matching the number on the gold ball (1 through 43 ) What is the probability of winning the minimum award? The probability of winning the minimum award is ◻ (Type an integer or a simplified fraction.)$

Probability with the $\newline$ Question $10$ , $11$ . $5$ . $9$ $\newline$ points $\newline$ Fundamental Countin $\newline$ Points: $0$ of $1$ $\newline$ Save $\newline$ Question list $\newline$ Question $6$ $\newline$ Question $7$ $\newline$ $x / 3$ Question $8$ $\newline$ * Question $9$ $\newline$ In a lottery, the top cash prize was $\$ 610$ million, going to three lucky winners. Players pick four different numbers from $1$ to $56$ and one number from $1$ to $43$ $\newline$ A player wins a minimum award of $\$ 300$ by correctly matching two numbers drawn from the white balls ( $1$ through $56$ ) and matching the number on the gold ball ( $1$ through $43$ ) What is the probability of winning the minimum award? $\newline$ The probability of winning the minimum award is $\square$ $\newline$ (Type an integer or a simplified fraction.)

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Math Problems

Algebra 2

Find probabilities using the addition rule

Full solution

Q. Probability with the

\newline

Question

10

11

5

9

\newline

points

\newline

Fundamental Countin

\newline

Points:

0

1

\newline

Save

\newline

Question list

\newline

Question

6

\newline

Question

7

\newline

x / 3

Question

8

\newline

* Question

9

\newline

In a lottery, the top cash prize was

\$ 610

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

1

56

and one number from

1

43

\newline

A player wins a minimum award of

\$ 300

by correctly matching two numbers drawn from the white balls (

1

through

56

) and matching the number on the gold ball (

1

through

43

) What is the probability of winning the minimum award?

\newline

The probability of winning the minimum award is

\square

\newline

(Type an integer or a simplified fraction.)

Calculate White Ball Combinations: To calculate the probability of winning the minimum award, we need to determine the probability of correctly matching two numbers from the white balls and one number from the gold ball.
Calculate Gold Ball Probability: First, let's calculate the probability of matching two numbers from the white balls. Since the order in which these numbers are drawn doesn't matter, we use combinations to calculate the number of ways to choose $2$ numbers out of $56$ . The number of combinations of $56$ taken $2$ at a time is calculated using the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items, and $k$ is the number of items to choose. $C(56, 2) = \frac{56!}{2!(56-2)!} = \frac{56!}{2!54!} = \frac{56 \times 55}{2 \times 1} = 1540$ .
Combine White Ball and Gold Ball Probabilities: Next, we calculate the probability of matching the number on the gold ball. There is only one correct gold ball number out of $43$ possible numbers. $\newline$ The probability of choosing the correct gold ball number is $1$ out of $43$ , which is $\frac{1}{43}$ .
Calculate Total Probability: Now, we need to combine the probabilities of both events happening together. Since these are independent events, we multiply their probabilities. $\newline$ The probability of matching two white balls is $2$ correct numbers out of the $4$ chosen, so we need to account for the fact that the player is choosing $4$ numbers in total. The probability of getting exactly $2$ correct out of $4$ chosen is $C(2, 2) \times C(54, 2) / C(56, 4)$ , where $C(54, 2)$ is the number of ways to choose the remaining $2$ numbers incorrectly from the $54$ non-winning numbers. $\newline$ $C(56, 4) = 56! / (4!(56-4)!) = (56 \times 55 \times 54 \times 53) / (4 \times 3 \times 2 \times 1) = 367290$ . $\newline$ $4$ $0$ . $\newline$ $4$ $1$ . $\newline$ So the probability of matching exactly two white balls is $4$ $2$ .
Calculate Total Probability: Now, we need to combine the probabilities of both events happening together. Since these are independent events, we multiply their probabilities. $\newline$ The probability of matching two white balls is $2$ correct numbers out of the $4$ chosen, so we need to account for the fact that the player is choosing $4$ numbers in total. The probability of getting exactly $2$ correct out of $4$ chosen is $C(2, 2) \times C(54, 2) / C(56, 4)$ , where $C(54, 2)$ is the number of ways to choose the remaining $2$ numbers incorrectly from the $54$ non-winning numbers. $\newline$ $C(56, 4) = 56! / (4!(56-4)!) = (56 \times 55 \times 54 \times 53) / (4 \times 3 \times 2 \times 1) = 367290$ . $\newline$ $C(54, 2) = 54! / (2!(54-2)!) = (54 \times 53) / (2 \times 1) = 1431$ . $\newline$ $C(2, 2) = 2! / (2!(2-2)!) = 1$ . $\newline$ So the probability of matching exactly two white balls is $(1 \times 1431) / 367290$ .Now we multiply the probability of matching two white balls with the probability of matching the gold ball to get the total probability of winning the minimum award. $\newline$ The probability of matching two white balls is $(1 \times 1431) / 367290$ , and the probability of matching the gold ball is $1/43$ . $\newline$ The total probability is $(1431 / 367290) \times (1 / 43)$ .
Calculate Total Probability: Now, we need to combine the probabilities of both events happening together. Since these are independent events, we multiply their probabilities. $\newline$ The probability of matching two white balls is $2$ correct numbers out of the $4$ chosen, so we need to account for the fact that the player is choosing $4$ numbers in total. The probability of getting exactly $2$ correct out of $4$ chosen is $C(2, 2) \times C(54, 2) / C(56, 4)$ , where $C(54, 2)$ is the number of ways to choose the remaining $2$ numbers incorrectly from the $54$ non-winning numbers. $\newline$ $C(56, 4) = 56! / (4!(56-4)!) = (56 \times 55 \times 54 \times 53) / (4 \times 3 \times 2 \times 1) = 367290.$ $\newline$ $4$ $0$ $\newline$ $4$ $1$ $\newline$ So the probability of matching exactly two white balls is $4$ $2$ .Now we multiply the probability of matching two white balls with the probability of matching the gold ball to get the total probability of winning the minimum award. $\newline$ The probability of matching two white balls is $4$ $2$ , and the probability of matching the gold ball is $4$ $4$ . $\newline$ The total probability is $4$ $5$ .Finally, we calculate the total probability by multiplying the two probabilities together. $\newline$ $4$ $6$