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.

There are 2 black queens in the deck, one of spades and one of clubs, so there is a 2/52 chance a black Queen will be drawn and 4/51 chance of drawing a 5 of any color, since the queen has already been removed from the deck. Thus:  2/52 * 4/51 = 8/2652 → 2/663.

1

To NOT choose a red or blue, leaves 6 black and 10 green to choose from. That leaves 16 marbles out of a total of 32 marbles, or a 1/2 chance.

We have blue, red, and green marbles in a bag. We need to consider all of the ways that we could draw at least one red. This means we can draw a red the first time, a red the second, or a red both times. These are the possible ways we could at least draw one red marble:

We can draw a red and then a blue.

We can draw a red and then a green.

We can draw a blue and then a red.

We can draw a green and then a red.

We could draw a red and then another red.

So we need to find the probability of each of these five scenarios. Then, we need to add these probabilities.

Let's look at the probability of the first scenario (red, then blue). The probability of drawing a red on the first time would be 4/10, because there are 10 marbles, and four are red. On the second draw, we don't put this marble back. This means we only have 9 marbles in the bag, and four of them are blue. Thus, the probability of the second draw being blue would be 4/9.

The probability of drawing a red and then a blue is equal to the product of these two events. Whenever we want to find the probability of one event AND another, we need to multiply. Thus, the probability of drawing the red AND then a blue would be (4/10)(4/9) = 16/90.

We can calculate the probability of the other four possibilities in a similar fashion.

The probability of drawing a red and then a green is (4/10)(2/9) = 8/90

The probability of drawing a blue then a red is (4/10)(4/9) = 16/90

The probability of drawing a green then a red is (2/10)(4/9) = 8/90

The probabilty of drawing a red then a red is (4/10)(3/9) = 12/90

To find the total probabilty, we need to add up the probabilities of the five different scenarios. Whenever we want to find the probability of one event OR another, we add. So the final probablity is

16/90 + 8/90 + 16/90 + 8/90 + 12/90 = 60/90 = 2/3.

The probability of drawing at least one red is 2/3.

Note that the choosing of a captain from each homeroom is independent of the other, such that the events are independent. This means that the probabilities of each event would be multiplied.

Find the probability that girls are selected for both homerooms. As this is the only possibility in which a boy is NOT selected, the answer is 1 minus this probability. 

There are 9 students in Homeroom A. Of them, 5 are girls.

There are 10 students in Homeroom B. Of them, 7 are girls.

First, observe that multiples of 5 are all those numbers that have a 0 or a 5 as their final digit (5, 10, 15, 20, 25, etc.), so the question is asking how many 4-digit integers under 7000 end in a 0 or a 5. Second, notice that the smallest 4-digit integer is 1,000. So, to rephrase the question, we want to find how many integers that are between 1000 and 6999 end in a 0 or a 5.

Writing the 4-digit integer as WXYZ, we see that there are six possibilties for W (1, 2, 3, 4, 5, and 6), ten possibilities for X (0–9), ten possibilities for Y (0–9), and two possibilities for Z (0 and 5). So there are 6 * 10 * 10 * 2 = 1200 total integers between 1000 and 6999 that end in a 0 or a 5. Therefore, there are 1200 four-digit integers that are multiples of 5 and less than 7,000.