There are 200 students in total, and 120 of them are male, so 80 must be female. We also know that there are 50 upper division students, and 40 of them are male, so 10 must be female. If 10 of 80 females are upper division, the other 70 females have to be lower division students, so the probability of choosing a lower division student, given the student is female, is 70/80 = 7/8.

We are given that the chosen dress has a defect, so we are looking at the 9 defective dresses. 3 of the defective dresses have rips, so the probability that a dress has a rip, given that is has a defect, is 3/9 = 1/3. 

Let's first find the probability of not getting a 1 on any of the three throws. Prob(no 1 on first throw) = 5/6. We need Prob(no 1 on first throw AND no 1 on second throw AND no 1 on third throw). "And" signifies multiplication in probability, so Prob(no 1 on all three throws) = 5/6 * 5/6 * 5/6 = 125/216. Now, to find the probability of getting a 1 on at least one throw, we simply take 1 – Prob(no 1 on all three throws) = 1 – 125/216 = 91/216.

We want either rainy, rainy, rainy or sunny, sunny, sunny. 

P(rain on Mon AND rain on Tues AND rain on Wed) = 3/5 * 3/4 * 1/3 = 9/60

P(sun on Mon AND sun on Tues AND sun on Wed) = 2/5 * 1/4 * 2/3 = 4/60

P(same weather) = P(all 3 rainy days OR all 3 sunny days)

                         = P(all 3 rainy days) + P(all 3 sunny days)

                         = 9/60 + 4/60 = 13/60.

The problem gives us extra information. We know 12 animals are brown, and there are 16 + 21 = 37 animals in total. Then the probability of choosing a brown animal is simply 12/37.

There are 52 cards in a standard deck. 13 of them are spades.

So there is a  chance of pulling out the first spade. That leaves 51 cards and 12 spades left. Now there is a chance to get a second spade.

 When you combine probabilities, multiply the two individual probabilites together.

There are 4 flips of the coin.  chance that the first is a heads, the second is a heads, the third is a tails, and the fourth is a tails. There are 6 permutations that give a combined 2 heads and 2 tails in whatever order.

There are 10 marbles in the bag. So, the first probability of pulling out the first red marble is ; the next is ; the final is .

Multiply the individual probabilities together to find the total outcome.

According to the table, 50 unvaccinated patients caught Rhinovirus.

18 + 4 + 5 + 4 + 19 = 50

250 unvaccinated patients caught Rhinovirus.

63 + 32 + 29 + 51 + 75 = 250

Next, we need to determine the percentage relationship between the number of unvaccinated versus vaccinated patients that caught the virus.

50/250 = 20%

100% – 20% = 80%

Because only 20% as many vaccinated patients exposed to the cold virus caught the cold, this reflects an 80% reduction in the likelihood of a patient to catch the cold after receiving the vaccine.

For the first card, the probability of choosing a diamond is . With no replacing, the probability of choosing another diamond is . Multilying these probabilities yields .

This reduces to .