P(boy) = 10/30 = 1/3

P(blue eyes) = 1/2

P(boy and blue eyes) = 5/30 = 1/6, because we are told that half (or 5) of the 10 boys have blue eyes

P(boy or blue eyes) = P(boy) + P(blue eyes) – P(boy and blue eyes)

= 1/3 + 1/2 – 1/6

= 2/3

As the options suggest, this problem requires us to know how probabilities work. The probability of a single event happening is equal to the number of ways it can happen divided by the total number of outcomes. Because this problem asks about 31 students, we have to raise our probability to the 31st power because it's the probability of 31 different events happening.

Now we can figure out the probability of one January birthday (31 days in January/365 days in a year) and the probability of all Saturday birthdays (52 Saturdays in a year/365 days in a year). So thus the probabilities that ALL of the students will have either type of birthday is:

(31/365)³¹ and

(52/365)³¹

Since the numbers are raised to the same power, we can simply look at the base to determine which is larger and which is smaller.  Since the probability of having a birthday on Saturday (roughly 1/7) is greater than having a birthday in January (roughly 1/12), then Quantity B is greater.

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.