Probability in Lottery Games: Winning Numbers from 1 to 65

Math Problem Statement

Read this:

A state lottery game requires that you pick 6 different numbers from 1 to

(a) Assuming that order is unimportant, what is the probability of picking all 6 numbers correctly to win the big prize?

(b) What is the probability of picking exactly

of the 6 numbers correctly?

Question content area bottom

Part 1

(a) Assuming that order is unimportant, what is the probability of picking all 6 numbers correctly to win the big prize?

Part 2

Let S be a sample space of equally likely outcomes, and let event E be a subset of S. Then the probability that event E occurs is given by the formula

P(E)equals=StartFraction n left parenthesis Upper E right parenthesis Over n left parenthesis Upper S right parenthesis EndFractionn(E)n(S).

Part 3

In this case, n(E) is the number of ways you can pick all 6 of the numbers correctly.

All 6 of the numbers can only be picked correctly 1 way.

Part 4

The sample space S is the set of all the ways you can pick 6 numbers. Since the numbers are selected without replacement and order does not matter,

n(S)equals=C(6565,6).

There are

82 comma 598 comma 88082,598,880

ways that 6 numbers can be picked.

Part 5

Find the probability of picking all 6 numbers correctly. Round to eleven decimal places and then convert the number to scientific notation.

P(E)

equals=

StartFraction n left parenthesis Upper E right parenthesis Over n left parenthesis Upper S right parenthesis EndFractionn(E)n(S)

equals=

StartFraction 1 Over 82 comma 598 comma 880 EndFraction182,598,880

equals=

1.211 times 10 Superscript negative 81.211×10−8

Part 6

Thus, the probability of picking all 6 numbers correctly is

1.211 times 10 Superscript negative 81.211×10−8.

Part 7

(b) What is the probability of picking exactly

of the 6 numbers correctly?

Part 8

Let

n(E)equals="exactly

of the 6 numbers drawn were selected." The outcomes in E contain any 6 numbers,

of which are from the 6 numbers drawn and the rest are not. Find the number of ways to select

numbers from the 6 numbers drawn.

Upper C left parenthesis 6 comma 5 right parenthesisC(6,5)

Part 9

From the total

6565

numbers, there are only six matching the numbers drawn on the ticket. Hence, find the quantity of numbers that do not match the numbers drawn.

6565minus−6equals=5959

Part 10

Find the number of ways to select the remaining

numbernothing

from the

5959

numbers that are not drawn on the ticket.

Upper C left parenthesis 59 comma 1 right parenthesisC(59,1)

Part 11

Thus, the total number of outcomes containing any six numbers,

of which are from the six numbers drawn on the ticket and the rest are not, can be found using the multiplication principle and is equal to

Upper C left parenthesis 6 comma 5 right parenthesisC(6,5)times•Upper C left parenthesis 59 comma 1 right parenthesisC(59,1).

Find P(E). Round to nine decimal places and then convert the number to scientific notation.

P(E)

equals=StartFraction n left parenthesis Upper E right parenthesis Over n left parenthesis s right parenthesis EndFractionn(E)n(s)

equals=StartFraction Upper C left parenthesis 6 comma 5 right parenthesis times Upper C left parenthesis 59 comma 1 right parenthesis Over 82 comma 598 comma 880 EndFractionC(6,5)•C(59,1)82,598,880

equals=4.286 times 10 Superscript negative 64.286×10−6

Part 12

Thus, the probability of picking exactly

of the 6 numbers drawn is

4.286 times 10 Superscript negative 64.286×10−6.

Solution

Let's break down the calculations for both parts of the problem:

Part (a): Probability of Picking All 6 Numbers Correctly

Understanding the Events:
- You have to choose 6 numbers correctly from a set of 65.
- The total number of ways to pick 6 numbers from 65 is given by the combination formula $C(n, k)$ , where $n$ is the total numbers to choose from, and $k$ is the number of choices.
Calculating the Sample Space: $n(S) = C(65, 6) = \frac{65!}{6!(65-6)!} = 82,598,880$
Calculating the Probability:
- There is exactly 1 way to pick all 6 winning numbers.
- Thus, $n(E) = 1$ .
Calculating the Probability: $P(E) = \frac{n(E)}{n(S)} = \frac{1}{82,598,880}$
Final Calculation:
- Rounding this to eleven decimal places gives: $P(E) \approx 1.211 \times 10^{-8}$

Part (b): Probability of Picking Exactly 5 of the 6 Numbers Correctly

Understanding the Scenario:
- You need to pick exactly 5 winning numbers and 1 losing number.
- You can choose 5 from the 6 winning numbers and 1 from the remaining 59 losing numbers.
Calculating $n(E)$ : $n(E) = C(6, 5) \times C(59, 1)$
- Calculate $C(6, 5)$ : $C(6, 5) = 6$
- Calculate $C(59, 1)$ : $C(59, 1) = 59$
Calculating Total Outcomes for Event E: $n(E) = 6 \times 59 = 354$
Calculating the Probability: $P(E) = \frac{n(E)}{n(S)} = \frac{354}{82,598,880}$
Final Calculation:
- Rounding this to nine decimal places gives: $P(E) \approx 4.286 \times 10^{-6}$

Summary of Results

Part (a): The probability of picking all 6 numbers correctly is $1.211 \times 10^{-8}$ .
Part (b): The probability of picking exactly 5 of the 6 numbers correctly is $4.286 \times 10^{-6}$ .

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

How would the probabilities change if the total numbers were different (e.g., picking from 70 instead of 65)?
What are the implications of these probabilities in real-world lottery scenarios?
How can we extend this problem to finding probabilities for picking fewer than 5 numbers correctly?
What are the different strategies players might use based on these probabilities?
How do combinations work in other probability scenarios, such as card games?

Tip: Understanding combinations and permutations is key in probability problems, especially in games of chance like lotteries!

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Math Problem Analysis

Mathematical Concepts

Probability
Combinatorics

Formulas

P(E) = n(E) / n(S)
C(n, k) = n! / (k!(n-k)!)

Theorems

Combination theorem

Suitable Grade Level

Grades 10-12

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