We have looked at the probability that two non-mutually exclusive events A or B can happen in this learning experience.

We will take it one step further and look at the probability that event A or B or C will happen.

The formula is:
P(A or B or C) = P(A) + P(B) + P(C)- P(what a and B has in common) - P(B and C have in common) - P(A and C have in common) + P(A,b,C have in common)
Use the formula to answer the following problem.

The numbers 1 to 40 are all written on slips of paper and placed into a bag. Find the probability to 3 decimal places of drawing a composite number above 20 or multiples of 4 or multiples of 3.

To find the probability of drawing a composite number above 20 or multiples of 4 or multiples of 3, we will use the formula P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C) + P(A, B, C).

Let's break down the problem:

A: Drawing a composite number above 20
B: Drawing a multiple of 4
C: Drawing a multiple of 3

Now we need to find the individual probabilities for each event:

P(A): To find the probability of drawing a composite number above 20, we need to determine how many numbers meet this condition. For this, we count the number of composite numbers above 20, which are 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, and 39. There are 15 such numbers. Since there are a total of 40 numbers (1 to 40), the probability of drawing a composite number above 20 is P(A) = 15/40.

P(B): To find the probability of drawing a multiple of 4, we count the numbers divisible by 4 between 1 and 40, which are 4, 8, 12, 16, 20, 24, 28, 32, 36, and 40. There are 10 such numbers, so P(B) = 10/40.

P(C): To find the probability of drawing a multiple of 3, we count the numbers divisible by 3 between 1 and 40, which are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, and 39. There are 13 such numbers, so P(C) = 13/40.

Now let's find the probabilities of the intersections:

P(A and B): We need to find the numbers that are both composite numbers above 20 and multiples of 4. From the list above, we see that 24 is the only number that satisfies both conditions. So, P(A and B) = 1/40.

P(B and C): We need to find the numbers that are both multiples of 4 and multiples of 3. From the list above, we see that 12, 24, and 36 satisfy both conditions. So, P(B and C) = 3/40.

P(A and C): We need to find the numbers that are both composite numbers above 20 and multiples of 3. From the list above, we see that 24 and 36 satisfy both conditions. So, P(A and C) = 2/40.

P(A, B, C): We need to find the numbers that satisfy all three conditions. From the list above, we find that only 24 satisfies all three conditions. So, P(A, B, C) = 1/40.

Now, plugging the values into the formula:

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C) + P(A, B, C)
P(A or B or C) = 15/40 + 10/40 + 13/40 - 1/40 - 3/40 - 2/40 + 1/40

Calculating the result:

P(A or B or C) ≈ 0.475

Therefore, the probability, to three decimal places, of drawing a composite number above 20 or multiples of 4 or multiples of 3 is approximately 0.475.