QuestionA bag has 20 papers numbered 1 to 20. Find the probability that 3 drawn numbers are even: (a) with replacement, (b) without replacement.

Studdy Solution

STEP 1

Assumptions1. The bag contains20 pieces of paper numbered from1 to20. We are drawing three numbers from the bag3. We are interested in the probability that all three numbers are even4. In scenario (a), each number is replaced after it is drawn5. In scenario (b), each number is not replaced after it is drawn

STEP 2

First, we need to find the total number of possible outcomes. In this case, since we are drawing three numbers from a set of20, the total number of possible outcomes is 20^ in scenario (a) and $\binom{20}{}$ in scenario (b).

STEP 3

Next, we need to find the total number of favorable outcomes. In this case, a favorable outcome is one in which all three numbers drawn are even. Since half of the numbers from1 to20 are even, there are10 even numbers. Therefore, the total number of favorable outcomes is $10^3$ in scenario (a) and $\binom{10}{3}$ in scenario (b).

STEP 4

Now, we can calculate the probability of drawing three even numbers in scenario (a). The probability is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$(A) = \frac{avorable\, outcomes\, (A)}{Total\, outcomes\, (A)}$

STEP 5

Plug in the values for the number of favorable outcomes and the total number of possible outcomes in scenario (a) to calculate the probability.
$(A) = \frac{10^3}{20^3}$

STEP 6

Calculate the probability of drawing three even numbers in scenario (a).
$(A) = \frac{10^3}{20^3} = \frac{1}{8}$

STEP 7

Now, we can calculate the probability of drawing three even numbers in scenario (b). The probability is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$(B) = \frac{avorable\, outcomes\, (B)}{Total\, outcomes\, (B)}$

STEP 8

Plug in the values for the number of favorable outcomes and the total number of possible outcomes in scenario (b) to calculate the probability.
$(B) = \frac{\binom{10}{3}}{\binom{20}{3}}$

STEP 9

Calculate the probability of drawing three even numbers in scenario (b).
$(B) = \frac{\binom{}{3}}{\binom{20}{3}} \approx.15$ The probability that all three numbers are even when you replace each number after it is drawn is $\frac{}{8}$ , and when you do not replace each number after it is drawn is approximately.15.

Was this helpful?

QuestionA bag has 20 papers numbered 1 to 20. Find the probability that 3 drawn numbers are even: (a) with replacement, (b) without replacement.

Studdy Solution

STEP 1

STEP 2

First, we need to find the total number of possible outcomes. In this case, since we are drawing three numbers from a set of20, the total number of possible outcomes is 20^ in scenario (a) and $\binom{20}{}$ in scenario (b).

STEP 3

STEP 4

Now, we can calculate the probability of drawing three even numbers in scenario (a). The probability is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$(A) = \frac{avorable\, outcomes\, (A)}{Total\, outcomes\, (A)}$

STEP 5

Plug in the values for the number of favorable outcomes and the total number of possible outcomes in scenario (a) to calculate the probability.
$(A) = \frac{10^3}{20^3}$

STEP 6

Calculate the probability of drawing three even numbers in scenario (a).
$(A) = \frac{10^3}{20^3} = \frac{1}{8}$

STEP 7

Now, we can calculate the probability of drawing three even numbers in scenario (b). The probability is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$(B) = \frac{avorable\, outcomes\, (B)}{Total\, outcomes\, (B)}$

STEP 8

Plug in the values for the number of favorable outcomes and the total number of possible outcomes in scenario (b) to calculate the probability.
$(B) = \frac{\binom{10}{3}}{\binom{20}{3}}$

STEP 9

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionA bag has 20 papers numbered 1 to 20. Find the probability that 3 drawn numbers are even: (a) with replacement, (b) without replacement.

Studdy Solution

STEP 1

STEP 2

First, we need to find the total number of possible outcomes. In this case, since we are drawing three numbers from a set of20, the total number of possible outcomes is 20^ in scenario (a) and (20)\binom{20}{}(20​) in scenario (b).

STEP 3

STEP 4

STEP 5

Plug in the values for the number of favorable outcomes and the total number of possible outcomes in scenario (a) to calculate the probability.(A)=103203(A) = \frac{10^3}{20^3}(A)=203103​

STEP 6

Calculate the probability of drawing three even numbers in scenario (a).(A)=103203=18(A) = \frac{10^3}{20^3} = \frac{1}{8}(A)=203103​=81​

STEP 7

STEP 8

Plug in the values for the number of favorable outcomes and the total number of possible outcomes in scenario (b) to calculate the probability.(B)=(103)(203)(B) = \frac{\binom{10}{3}}{\binom{20}{3}}(B)=(320​)(310​)​

STEP 9

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

First, we need to find the total number of possible outcomes. In this case, since we are drawing three numbers from a set of20, the total number of possible outcomes is 20^ in scenario (a) and $\binom{20}{}$ in scenario (b).

Plug in the values for the number of favorable outcomes and the total number of possible outcomes in scenario (a) to calculate the probability.
$(A) = \frac{10^3}{20^3}$

Calculate the probability of drawing three even numbers in scenario (a).
$(A) = \frac{10^3}{20^3} = \frac{1}{8}$

Plug in the values for the number of favorable outcomes and the total number of possible outcomes in scenario (b) to calculate the probability.
$(B) = \frac{\binom{10}{3}}{\binom{20}{3}}$