40 students live in a dormitory. 5 have swine flu. If 4 students are selected at random, what is the probability that 1 or 2 of the 4 have swine flu?

Probability of 0: (35/40)(34/39)(33/38)(32/37) = 0.5729

Probability of 4: (5/40)*(4/39)*(3/38)*(2/37) = 120/2,193,360 = 0.000055

Probability of 1:
4*(35/40)*(34/39)*(33/38)*(5/37)= 0.3581

Probability of 2:
(4!/(2!*2!))(35/40)*(34/39)*(5/38)*(4*37)= 0.0651

To calculate the probability that 1 or 2 out of the 4 selected students have swine flu, we need to consider two cases: one where exactly 1 student has swine flu, and another where exactly 2 students have swine flu.

Let's break it down step by step:

Step 1: Calculate the total number of ways to select 4 students from a group of 40. This can be done using the combination formula, expressed as C(40,4), which is calculated as:

C(40, 4) = 40! / (4! * (40-4)!),
where "!" represents the factorial function.

Simplifying the equation:
C(40, 4) = (40 * 39 * 38 * 37) / (4 * 3 * 2 * 1) = 91390.

So, there are 91390 different ways to select 4 students from a group of 40.

Step 2: Calculate the number of ways to select 1 student with swine flu and 3 students without swine flu.

There are 5 students with swine flu, so we need to choose 1 from those 5 students. We multiply this by the number of ways to choose the remaining 3 students from the 35 students without swine flu. This can be calculated as:

C(5, 1) * C(35, 3) = 5 * (35! / (3! * (35-3)!)) = 5 * 6545 = 32725.

So, there are 32725 different ways to select 1 student with swine flu and 3 students without swine flu.

Step 3: Calculate the number of ways to select 2 students with swine flu and 2 students without swine flu.

There are 5 students with swine flu, so we need to choose 2 from those 5 students. We multiply this by the number of ways to choose the remaining 2 students from the 35 students without swine flu. This can be calculated as:

C(5, 2) * C(35, 2) = (5! / (2! * (5-2)!)) * (35! / (2! * (35-2)!)) = 10 * 595 = 5950.

So, there are 5950 different ways to select 2 students with swine flu and 2 students without swine flu.

Step 4: Add the number of ways calculated in steps 2 and 3 to get the total number of ways to select 1 or 2 students with swine flu:

Total number of cases = 32725 + 5950 = 38675.

Step 5: Calculate the probability by dividing the total number of cases (from step 4) by the total number of possible selections (from step 1):

Probability = Total number of cases / Total possible selections = 38675 / 91390.

Calculating this value:
Probability ≈ 0.423.

Therefore, the probability that 1 or 2 out of the 4 selected students have swine flu is approximately 0.423 or 42.3%.