Question

In a class of 27 students, 17 are dual-enrolled and 10 are not dual-enrolled. Suppose two students are randomly selected from the class (without replacement). Calculate the following probabilities. Round solutions to three decimal places, if necessary.

Answer 1

The probability that the first student is dual-enrolled and the second student is not dual-enrolled is approximately 0.207 (rounded to three decimal places).

To calculate the probabilities, we can use the concepts of conditional probability and combinations (since we are selecting students without replacement). Let's calculate each probability step by step:

Step 1: Calculate the probability that the first student is dual-enrolled (DE).

Given:

- Total students in the class (\( N \)) = 27

- Dual-enrolled students (\( DE \)) = 17

`\[ P(\text{1st DE}) = (DE)/(N) = (17)/(27) \]`

Step 2: Calculate the probability that the second student is dual-enrolled given that the first student was dual-enrolled.

`\[ P(\text2nd DE ) = (DE - 1)/(N - 1) = (17 - 1)/(27 - 1) = (16)/(26) \]`

Step 3: Calculate the probability that the first student is dual-enrolled and the second student is not dual-enrolled.

`\[ P(\text{1st DE and 2nd Not DE}) = 1 - \left((17)/(27) * (16)/(26)\right) \]`

Now, let's calculate \( P(\text{1st DE and 2nd Not DE}) \):

`\[ P(\text{1st DE and 2nd Not DE}) = 1 - \left((17)/(27) * (16)/(26)\right) = 1 - (16 \cdot 17)/(26 \cdot 27) \]`

Calculating this:

`\[ P(\text{1st DE and 2nd Not DE}) \approx 0.207 \]`

So, the probability that the first student is dual-enrolled and the second student is not dual-enrolled is approximately 0.207 (rounded to three decimal places).

Answer 2

Explanation:

a probability is always the ratio

desired cases / totally possible cases.

if 2 non-overlapping events should occur, the probability of both events happening is the product of both individual probabilities.

e.g. the probability to roll two 6 with 2 dice (or in 2 consecutive rolls with 1 die) is 1/6 × 1/6 = 1/36

if 1 of 2 possible non-overlapping events should occur, the probability of one of the events happening is the sum of the individual probabilities.

e.g. the probability to roll a 1 or a 2 with a die is

1/6 + 1/6 = 2/6 = 1/3

both students are dual-enrolled.

in our case, for the first selection we have the totally possible cases of 27.

the desired cases (dual-enrolled) are 17.

so, the probabilty to select a dual-enrolled student is

17/27

for the second selection (without replacement of the first pull) the totally possible cases are now 26 (because we pulled one student from the general pool of candidates in the first selection).

the desired cases are now 16 (because for our case we pulled a dual-enrolled student from the general pool).

the probabilty to select a dual-enrolled student in the second selection is

16/26 = 8/13

the overall probability to select two dual-enrolled students in 2 selections is

17/27 × 8/13 = 136/351 = 0.387464387... ≈ 0.387

the first student is dual-enrolled, the second is not.

for the first selection the totally possible cases are 27.

the desired cases are again 17.

the probability of the first dejection is again

17/27

for the second selection the totally possible cases are 26 (see above).

the desired cases are still 10.

the probably for this second selection is

10/26 = 5/13

the overall probability to select first a dual-enrolled student and then a not-dual-enrolled student is

17/27 × 5/13 = 85/351 = 0.242165242... ≈ 0.242