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.

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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.

First, find the amount of time that the light is yellow,  which is 50 – 20 – 25 = 5 seconds out of a 50 second cycle. The period of 100 seconds is irrelevant because the probability of getting a yellow light would be the same for any complete period or number of complete periods. So the probability of getting a yellow light would be 5/50 = 1/10.

There is a 1/3 chance that a 2 or 6 will come up for any given roll.  The probability that one will come up in 30 rolls is (1/3) * 30 = 10

A 2 or 6 will, on average, appear 10 times in 30 rolls.

The probability a heads will show on any flip is 1/2.  To find the probability of 4 consecutive heads, multiply the probability of each individual flip together.

1/2 * 1/2 * 1/2 * 1/2 = 1/16

Our strategy is as follows: We must determine the number of possible entrees, the number of possible combinations of two side dishes, and the number of desserts. Then we must multiply these four numbers together to get the number of all of the possible meals.

Let's label the two different entrees E₁ and E₂ The customer can choose one of these, so he has two different options for his entree.

Let's label the four different side dishes S₁, S₂, S₃, and S₄. The customer can only choose two of these four dishes, and the order in which he chooses doesn't matter. For example, if he were to choose mashed potatoes and broccoili, he would end up with the same meal as if he were to choose broccoli and then mashed potatoes. So we need to consider all of the different pairs that can be selected from these four side dishes. Here are all of the possible combinations: 

S₁ and S₂

S₁ and S₃

S₁ and S₄

S₂ and S₃

S₂ and S₄

S₃ and S₄

This means that the customer has six different combinations of two side dishes.

Finally, the customer can choose one dessert from three different choices. This means there are three possibilities for dessert. 

So, the customer has two different options for entrees, six different options for the two side dishes, and three different choices for dessert. To find the total number of meals possible, we can multiply these three numbers together.

Number of meals = 2 x 6 x 3 = 36.

The answer is 36.

The number of possibilities in a joint event (that is, the entire four-digit integer) is the product of the number of possibilities of each individual event (the number of possibilities for each digit individually). If WXYZ represents a 4-digit integer, the problem tells us that there is only one possible value for W (which is 8) and only possible value for Z (which is 5). We also know that there are ten possibilities each for X and Y (they can be the digits 0-9). So the number of possible values for WXYZ is the product of the possible values for each digit, which is 1 * 10 * 10 * 1 = 100.