PSAT Math : Statistics

Study concepts, example questions & explanations for PSAT Math

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Example Questions

Example Question #191 : Data Analysis

A bag contains red marbles and blue marbles only. Let r equal the number of red marbles and b equal the number of blue marbles in the bag. Mark draws one marble from the bag and then a second one without replacing the first. Which of the following expressions is equivalent to the probability that Mark will draw a red marble and then a blue marble?

Possible Answers:

rb/(r+ b2)

rb/(r+ 2br + b2)

(r + b – 1)/(r+ 2br + b2)

rb/(r+ b+ 2br – r – b)

(rb – r – b)/(r+ b+ 2br – r – b)

Correct answer:

rb/(r+ b+ 2br – r – b)

Explanation:

Prob_part_1

Prob_part_2

Example Question #192 : Data Analysis

In a bag, there are 5 blue marbles, 3 red marbles, and 2 green marbles.  If three marbles are chosen consecutively at random without replacement, what is the probability that the color of the chosen marbles would be blue-red-green in that order?

Possible Answers:

1/3

1/24

3/100

1/300

1/1000

Correct answer:

1/24

Explanation:

The probability that a blue marble is chosen first is 5/10 because at the outset there are 5 blue marbles in a total of 10 marbles.  After one marble has been chosen, there are 9 total marbles remaining since there is no replacement.  The probability that a red marble is chosen in the second pick is 3/9 (because there are 3 red marbles out of a total of 9).  There are now 8 total marbles remaining and 2 green marbles. Thus, the probability of picking a green marble on the third pick is 2/8. 

 

Therefore, the probability of picking a blue-red-green marble outcome (in that order) is:

5/10 * 3/9 * 2/8 = 30/720 = 1/24

Example Question #193 : Data Analysis

A book shelf has 15 theology books, 20 philosophy books, and 5 history books. If someone were to draw two texts at random, what is the chance that they would draw at least one theology book?

Possible Answers:

None of the other answers

9/64

44/121

8/13

5/13

Correct answer:

8/13

Explanation:

Thinking through our data, we know that we are looking for the following combination of events, where T is a theology book, and < > represents a draw from the shelf:

1. <T><T>

2. <T><Non-T>

3. <Non-T><T>

To save us quite a bit of trouble, let us note that only one event is excluded:

<Non-T><Non-T

The easiest way to solve this would be to solve for the probability of this one case and subtract that from 1. This will give us the "remaining probability" that applies to the three cases that we want.

The <Non-T><Non-T> case would be calculated:

First draw: 25/40 = 5/8

Second draw: 24/39 = 8/13

Total probability: (5/8) * (8/13) = 5/13

The probability of our case is 1 – (5/13) = (13 – 5)/13 = 8/13.

Example Question #31 : Probability

A drawing is being held for concert tickets at a high school. There are 200 seniors, 150 juniors, 200 sophomores and 150 freshmen. Each senior's name is placed in the drawing 4 times, each junior's 3, sophomores 2, and freshmen one time. What is the probability that a senior's name will be chosen?

Possible Answers:

5/9

2/7

4/9

2/9

2/9

Correct answer:

4/9

Explanation:

You multiply each number of students in each class by how many times that class is entered into the drawing, then add up the totals. There will be a total of 1800 entrants and 800 will be seniors.

Example Question #31 : How To Find The Probability Of An Outcome

Set S = {0, 1, 5, 9}, and set T = {2, 3, 4, 7}. If one number is chosen randomly from S and another is chosen randomly from T, what is the probability that the sum of these two numbers will be prime?

Possible Answers:

7/16

5/8

9/16

1/2

1/4

Correct answer:

1/2

Explanation:

There are four possible numbers that can be chosen from S, and there are four that can be chosen from T. This means that there are 4 * 4, or 16, pairs of numbers that can be drawn from S and T. We need to find the sum of these pairs and determine if each sum is prime. Let's find the sum of the 16 pairs. Remember that a number is prime if it divisible only by itself and 1.

0 + 2 = 2, which is prime

0 + 3 = 3, prime

0 + 4 = 4, not prime

0 + 7 = 7, prime

1 + 2 = 3, prime

1 + 3 = 4, not

1 + 4 = 5, prime

1 + 7 = 8, not

5 + 2 = 7, prime

5 + 3 = 8, not

5 + 4 = 9, not

5 + 7 = 12, not

9 + 2=11, prime

9 + 3 = 12, not

9 + 4 = 13, prime

9 + 7 = 16, not

Of the sixteen pairs, 8 have a sum that equals a prime number. Thus, because each of these pairs has an equal chance of being drawn randomly, the probability that the sum will be prime is 8 out of 16, or 8/16 = 1/2.

The answer is 1/2.

Example Question #33 : Outcomes

Judy is practicing to be a magician and has an ordinary deck of 52 playing cards, a regular 6-sided die, and a fair coin. What is the probability that Judy rolls a 6 on the die, then flips the coin head face up, and then draws a spade from the deck of cards?

Possible Answers:

3/16

1/48

11/12

1/4

2/25

Correct answer:

1/48

Explanation:

We can find the individual probabilities of these three events occuring first. 

P(rolling a 6) = 1/6

P(head) = 1/2

P(spade) = 13/52 = 1/4

Now, to find the probability of rolling a 6 AND flipping a head AND drawing a spade, we must multiply the individual probabilities. So the answer is 1/6 * 1/2 * 1/4 = 1/48.

Example Question #32 : Probability

A bag of marbles has 7 yellow marbles, 5 red marbles, 3 blue marbles, and 6 white marbles. What is the probability of choosing a yellow marble, putting it back and choosing a blue marble, and then NOT putting the blue marble back and picking a white marble?

Possible Answers:

3/17

8/31

1/9

2/3

1/70

Correct answer:

1/70

Explanation:

There are a total of 7 + 5 + 3 + 6 = 21 marbles. The probability of picking a yellow marble is 7/21 = 1/3. Then we put it back and choose a blue marble with probability 3/21 = 1/7. We do NOT put this blue marble back, but then we grab for a white. The probability of picking a white is now 6/20 = 3/10, because now we are choosing from 20 marbles instead of 21. So putting it together, the probability of choosing a yellow marble, replacing it and then choosing a blue and a white, is 1/3 * 1/7 * 3/10 = 1/70. 

Example Question #41 : How To Find The Probability Of An Outcome

In a regular 52-card deck, what is the probability of drawing three aces in a row, with replacement?

Possible Answers:

4/52 * 4/52 * 4/52

1/52 + 1/52 + 1/52

1/52 * 1/51 * 1/50

13/52 * 13/51 * 13/50

1/52 * 1/52 * 1/52

Correct answer:

4/52 * 4/52 * 4/52

Explanation:

There are 4 aces in the 52 card deck, so the probability of drawing an ace is 4/52. Then we put this ace back in the deck and draw again. The probability of drawing an ace is again 4/52. Similarly, on the third draw, the probability of getting an ace is 4/52. So the probability of drawing an ace on the 1st draw AND the 2nd draw AND the 3rd draw is 4/52 * 4/52 * 4/52.

Example Question #41 : Probability

A bag contains four blue marbles, eight red marbles, and six orange marbles. If a marble is randomly selected from the bag, what is the probability that the marble will be either red or orange?

Possible Answers:

7/9

2/5

2/3

5/9

1/3

Correct answer:

7/9

Explanation:

The probability of an event is the ratio of the number of desired outcomes to the total number of possible outcomes. In this problem, the total number of outcomes is equal to the total number of marbles in the bag. There are four blue, eight red, and six orange marbles, so the total number of marbles is the sum of four, eight, and six, or eighteen.

We are asked to find the probability of choosing a marble that is either red or orange. This means we have to consider the number of marbles that are either red or orange. Because there are eight red and six orange, there are fourteen marbles that are either red or orange. 

The probability is thus fourteen out of eighteen, because there are fourteen red or orange marbles, out of a total of eighteen marbles. We will need to simplify the fraction 14/18.

Probability = 14/18 = 7/9

The answer is 7/9.

Example Question #42 : Probability

Two dice are rolled. Find the probability that the numbers sum to 4.

Possible Answers:

\frac{1}{6}

\frac{1}{4}

\frac{1}{12}

\frac{1}{15}

\frac{1}{3}

Correct answer:

\frac{1}{12}

Explanation:

The possible dice combinations that sum to 4 are \left \{ (1,3),(2,2),(3,1) \right \}.

The number of all possible dice combinations is 6\times 6=36. (6 numbers on each of the two dice.)

So the probability that the numbers sum to 4  =

\frac{3\ outcomes\ that\ sum\ to\ 4}{36\ possible\ outcomes}=\frac{3}{36}=\frac{1}{12}

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