Add up the total number of jellybeans, 20 + 45 + 30 + 5 = 100. 

Divide the number of watermelon jellybeans by the total: 20/100 and reduce the fraction to 1/5.

The probability of the point being inside the circle is the ratio of the area of the circle to the area of the square. If we suppose that the circle has radius r, then the square must have side 2r. The area of the circle is πr² and the area of the square is 〖(2r)〗²=〖4r〗², so the proportion of the areas is (πr²)/〖4r〗²  =π/4.

The bowl has 54 marbles, half green and half blue. This gives us 27 green and 27 blue marbles:

27 G / 27 B

John then takes 3 green and 6 blue from the bowl. This leaves the bowl with:

24 G / 21 B

If there are going to be more blue than green marbles after John's 13 marbles, he has to take at least 4 more green marbles than blue marbles, because right now there are 3 less blue marbles. Therefore, we need to take at least 9 green marbles, which would mean 4 or less of the marbles would be blue (8 green and 5 blue would leave us with equal green and equal blue marbles, so it would have to be more than 8 green marbles, which gives us 9 green marbles).

We can also solve this as an inequality. You take the difference in marbles, which is 3, which means you need the difference in green and blue marbles to be greater than 3, or at least 4. You have b + g = 13 and g - b > 3, where b and g are positive integers.

b + g = 13 (Subtract g on both sides of the equation)

b = 13 - g

g - b > 3 (Substitute above equation)

g - (13 - g) > 3 (Distribute negative sign in parentheses)

g - 13 + g > 3 (Add both g variables)

2g - 13 > 3 (Add 13 to both sides of the inequality)

2g > 16 (Divide both sides of the inequality by 2)

g > 8 so g has to be 9 or greater.

If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?

Here we have 5 possible choices for x and 4 possible choices for y, giving us 5 * 4 = 20 possible outcomes.

We know that odd times odd = odd; even times even = even; and even times odd = even. Thus we need all of the outcomes where x and y are odd. We have 3 possibilities of odd numbers for x, and 3 possibilities of odd numbers for y, so we will have 9 outcomes of our total 20 outcomes where xy is odd, giving us a probability of 9/20.

There are 18 marbles in total. One of them is removed so now there are 17 marbles. This is our denominator. All of the original white marbles are still in the bag so there is a 4 out of 17 or 4/17 chance that the next marble taken out of the bag will be white.

Michael can toss either one head, two heads, or three heads.

If Michael tosses one head, then it could be on either the first, second, or third toss. We could model it like this, where H represents heads and T represents tails.

HTT, THT, or TTH

If Michael tosses two heads, then there are three possible combinations:

HHT, HTH, or THH

If Michael tosses three heads, then there is only one possible combination:

HHH

Thus, there are seven ways that Michael can toss at least one head. We must find the probability of each of these ways and then add them together.

The probability of rolling a head is ½ and the probability of rolling a tail is ½. Because each coin toss is independent, we can multiply the probabilities together.

For example, the probability of the combination HTT is (1/2)(1/2)(1/2) = 1/8

Probability of HTT = 1/8

Probability of THT = (1/2)(1/2)(1/2) = 1/8

Probability of TTH = (1/2)(1/2)(1/2) = 1/8

Probability of HHT = 1/8

Probability of HTH = 1/8

Probability of THH = 1/8

Probability of HHH = 1/8

So, there are seven possible ways that Michael can toss at least one head. The probability of each of these seven ways is equal to 1/8. Thus, the total probability of all seven events is 7/8.

ALTERNATE SOLUTION:

Michael can toss at least one head, or he can toss zero heads. The sum of these two probabilities must equal one, because they represent all of the ways that Michael could toss the coins. He could either toss at least on head, or he could toss no heads at all.

Probability of tossing at least one head + probability of tossing no heads = 1

The probability of tossing no heads is only possible with the combination TTT. The probability of tossing three tails is equal to (1/2)(1/2)(1/2) = 1/8

Probability of tossing at least one head + 1/8 = 1

Probability of tossing at least one head = 1 – 1/8 = 7/8 .

Probability of each event = (# green marbles + # blue marbles)/ Total # of Marbles

P1 = (15 + 25) / 125 = 40 / 125

Second event assumes a blue or green was chosen for first event so there is one fewer marble on top and also one fewer marble in the total number of marbles.

P2 = (14 + 25) / 124 = 39 / 124

Third event assumes a blue or green was chosen for first and second events so there are two fewer marbles on top and also two fewer marbles in the total number of marbles.

P3 = (13 + 25) / 124 = 38 / 123

Probability for multiple events = P1 x P2 x P3

(40 / 125) * (39 / 124) * (38 / 123)

( 40 * 39 * 38) / (125 * 124 * 123 ) = 59280 / 1906500 = 0.031

There are 36 possible outcomes of the additive dice roll. The way to roll a sum of 9 is 6 (and vice versa) and 3 or 5 and 4 (and vice versa). This is possible 4 of the 36 times, giving a probability the sum of the two dice rolled of 4/36 or 1/9.

There are 20 total marbles. Selecting the first green marble has a 6/20 chance, the second green marble has a 5/19 chance. This gives a total chance of 30/380, or a 3/38 chance. 

The first thing to do here is find the total number of students in the contest. Seniors = 300 * 5 = 1500, Juniors = 200 * 3 = 600, Sophomores = 200, and Freshmen = 100. So adding all these up you get a total of 2400 names in the hat. Out of these 2400 names, 600 of them are Juniors. So the probability of choosing a Junior's name is 600/2400 = 1/4.