### All SAT Math Resources

## Example Questions

### Example Question #3091 : Sat Mathematics

Michelle is randomly drawing cards from a deck of of 52 cards. What is the chance she will draw a black queen followed by a 5 of any color, without replacing the cards?

**Possible Answers:**

4/169

8/663

2/663

2/169

2/2652

**Correct answer:**

2/663

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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### Example Question #3092 : Sat Mathematics

Zack has 10 green, 14 red, 2 blue, and 6 black marbles in a bag. What is the probability that Zack will not randomly pick a red or blue marble from the bag?

**Possible Answers:**

1/3

3/16

1/2

15/32

5/16

**Correct answer:**

1/2

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.

### Example Question #3093 : Sat Mathematics

A bag contains four blue marbles, four red marbles, and two green marbles in a bag. If one marble is drawn, and then a second is drawn without replacement, what is the probability that at least one of the two marbles will be red?

**Possible Answers:**

**Correct answer:**2/3

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.

### Example Question #3094 : Sat Mathematics

There are 5 girls and 4 boys in Homeroom A, and 3 boys and 7 girls in Homeroom B. If each homeroom selects one captain at random from among its students, what is the probability that a boy is selected in at least one of the homerooms?

**Possible Answers:**

7/18

11/10

67/90

11/18

12/19

**Correct answer:**

11/18

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.

Probability of 2 girls = P(girl for A) x P(girl for B)

= (5/9) x (7/10) = 35/90 = 7/18

Probability of at least one boy = 1 - 7/18 = 11/18

### Example Question #3095 : Sat Mathematics

How many positive four-digit integers are multiples of 5 and less than 7,000?

**Possible Answers:**

800

1850

1600

1000

1200

**Correct answer:**

1200

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.

### Example Question #11 : How To Find The Probability Of An Outcome

A street light in Anytown, USA, completes a cycle from red to green to yellow, and back to red, in 50 seconds. During this time, the light will be red for 20 seconds, green for 25 seconds, and yellow for the remaining time. Over a period of 100 seconds, what is the probability that the light will be yellow at a randomly chosen time?

**Possible Answers:**

1/5

1/50

1/10

1/20

1/25

**Correct answer:**

1/10

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.

### Example Question #12 : How To Find The Probability Of An Outcome

A fair die is rolled 30 times. How many times would you expect a 2 or a 6 to be rolled?

**Possible Answers:**

12

None of the other answers

5

10

15

**Correct answer:**

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.

### Example Question #11 : How To Find The Probability Of An Outcome

A coin is flipped 4 times. What is the probability of getting 4 heads in a row?

**Possible Answers:**

**Correct answer:**1/16

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

### Example Question #12 : How To Find The Probability Of An Outcome

A restaurant offers two different entrees, four different side dishes, and three desserts. If a customer wants to order a meal special that consists of one entree, two different side dishes, and one dessert, how many different meals are possible?

**Possible Answers:**

**Correct answer:**36

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 E_{1} and E_{2} The customer can choose one of these, so he has two different options for his entree.

Let's label the four different side dishes S_{1}, S_{2}, S_{3}, and S_{4}. 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:

S_{1 }and S_{2}

S_{1 }and S_{3}

S_{1 }and S_{4}

S_{2} and S_{3}

S_{2} and S_{4}

S_{3} and S_{4}

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.

### Example Question #13 : How To Find The Probability Of An Outcome

How many positive four-digit integers have the thousands digit equal to 8 and the units digit (ones digit) equal to 5?

**Possible Answers:**

20

10

100

99

200

**Correct answer:**

100

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.