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PSAT Math : Statistics

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Example Questions

Example Question #42 : Probability

Jackie is a contestant on a gameshow. She has to pick marbles out of a big bag to win various amounts of money. The bag contains a total of 200 marbles.

There are 100 red marbles worth $10 each, 50 blue marbles worth $20 each, 30 green marbles worth $50 each, 15 white marbles worth $100 each and 5 black marbles worth $1000 each. If she picks once, what percent chance does she have of picking a $1000 marble?

Possible Answers:

2.5%

None of the other answers

.05%

Correct answer:

2.5%

Explanation:

There are 5 black marbles worth $1000 dollars out of the total of 200 marbles. $\dpi{100} \small 5\div 200=.025$ or 2.5%

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Example Question #44 : Probability

What is the probability that you will pull out 3 diamonds in a standard deck without replacement.

Possible Answers:

$\dpi{100} \small \frac{1}{4}$

$\dpi{100} \small \frac{1}{1250}$

$\dpi{100} \small \frac{11}{850}$

$\dpi{100} \small \frac{1}{64}$

$\dpi{100} \small \frac{3}{52}$

Correct answer:

$\dpi{100} \small \frac{11}{850}$

Explanation:

There are 13 diamonds in a standard deck of 52 cards. So, you have a 13/52 (1/4) chance of getting a diamond; then a 12/51 (4/17) chance of pulling the next diamond; last, there is a 11/50 chance of getting the third diamond. When you combine probabilities, you multiply the individual probabilities together

$\dpi{100} \small \frac{1}{4}\times \frac{4}{17}\times \frac{11}{50}=\frac{11}{850}$

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Example Question #43 : Probability

On Halloween there is a bowl of candy. 33 pieces are gum drops, 24 are candy corn, 15 are suckers and 28 are hard candies. Without looking in the bowl, what are the chances that you pull a gum drop?

Possible Answers:

$\dpi{100} \small 0.38$

$\dpi{100} \small 0.50$

$\dpi{100} \small 0.49$

$\dpi{100} \small 0.33$

$\dpi{100} \small 0.67$

Correct answer:

$\dpi{100} \small 0.33$

Explanation:

Probability = # Gum Drops / # Total Number of Candies

$\dpi{100} \small \frac{33}{(33+24+15+28)}=\frac{33}{100}=0.33$

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Example Question #163 : Data Analysis

What is the probability of pulling out 4 diamonds in a standard deck of cards, without replacement?

Possible Answers:

$\frac{1}{13}$

$\frac{1}{256}$

$\frac{1}{4}$

$\frac{11}{4165}$

$\frac{1}{1013}$

Correct answer:

$\frac{11}{4165}$

Explanation:

There are 52 cards in a standard deck; 13 of them are diamonds. This means that there is a $\frac{13}{52}$ chance that the first card you pick will be a diamond. The probability of pulling another diamond as your next card is $\frac{12}{51}$ ; next diamond $\frac{11}{50}$ ; last diamond $\frac{10}{49}$ .

In order to get one probability, mulitply the individual probabilities together.

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Example Question #44 : Probability

A coporation is deciding who should sit on its leadership committee. If twenty people have applied, and the corporation will choose six applicants to sit on the committee, how many committees are possible?

Possible Answers:

$\dpi{100} 1,860,480$

$720$

$\dpi{100} 240,240$

$\dpi{100} 38,760$

$\dpi{100} 116,280$

Correct answer:

$\dpi{100} 38,760$

Explanation:

This question requires us to make use of the combination formula. In general, if we have $\dpi{100} n$ total objects or things to choose from, the number of different groups of size $\dpi{100} r$ that we can make is equal to the following:

$\frac{n!}{(r!)(n-r)!}$

where the ! symbol denotes a factorial. In general, $\dpi{100} k!=k(k-1)(k-2)...(3)(2)(1)$ .

For example, $\dpi{100} 6!=6\times 5\times 4\times 3\times 2\times 1=720$ .

We have 20 people from whom to select, and each group is going to have 6 people. Therefore, we will let $\dpi{100} n=20$ and $\dpi{100} r=6$ .

number of combinations = $\frac{20!}{(6!)(20-6)!}=\frac{20!}{(6!)(14)!}$

$\frac{(20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot \cdot \cdot 3\cdot 2\cdot 1)}{(6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)(14\cdot \cdot \cdot 3\cdot 2\cdot 1)}$

= $\frac{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$

$=38760$

The answer is $\dpi{100} 38,760$ .

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Example Question #44 : Probability

What is the probability that two dice rolled will add up to 5?

Possible Answers:

$\frac{1}{18}$

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{9}$

$\frac{2}{3}$

Correct answer:

$\frac{1}{9}$

Explanation:

There are 6 probabilities for each dice and there are 2 dice. So, you would multiply the individual probabilities together and get 36 possible outcomes.

In order for 2 dice to add to a sum of 5, you would need to roll:

1 4; 2 3; 3 2; 4 1 [each dice can roll as either number]

That is a total of 4 outcomes

$\frac{4}{36}=\frac{1}{9}$

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Example Question #45 : Probability

A bag containing only red, green and blue marbles has 10 red, 20 green, and 20 blue marbles. If Laura reaches in and removes a single marble at random, what is the probability she draws a green marble?

Possible Answers:

$\small \frac{1}{2}$

$\small \frac{1}{10}$

$\small \frac{1}{5}$

$\small \frac{2}{5}$

$\small \frac{3}{5}$

Correct answer:

$\small \frac{2}{5}$

Explanation:

Probability = the number of desired outcomes divided by the number of total possible outsomes. 20 marbles are green out of a total of 50. Therefore:

$\small \frac{20}{50}=\frac{2}{5}$

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Example Question #41 : Probability

A deck of 52 cards is divided equally into four suits: hearts, diamonds, clubs and spades, and two cards are drawn at random. The first card drawn is a heart. What is the probability that the second card drawn is also a heart?

Possible Answers:

$\small \frac{1}{16}$

$\small \frac{1}{3}$

$\small \frac{13}{51}$

$\small \frac{1}{4}$

$\small \frac{4}{17}$

Correct answer:

$\small \frac{4}{17}$

Explanation:

There are 13 hearts in the deck of 52 cards. The first card is already assumed to be a heart, but by removing it the numbers change. When drawing the second card, there are now 12 hearts out of 51 total cards.

$\small \small \frac{12}{51}\,\: \textup{simplify:}\: \,\frac{4}{17}$

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Example Question #51 : Probability

A game is played where players take turns rolling two six-sided dice, each with sides numbered 1-6. The numbers on the two dice are added together for each player's score. If the first player rolls a sum of 10, what is the probability that the second player rolls a HIGHER total?

Possible Answers:

$\small \frac{1}{18}$

$\small \frac{1}{10}$

$\small \frac{1}{12}$

$\small \frac{2}{6}$

$\small \frac{1}{4}$

Correct answer:

$\small \frac{1}{12}$

Explanation:

A roll of two 6-sided dice yields 36 total possible outcomes $\left ( 6\times 6 \right )$ . There are three possible ways to roll a sum higher than 10:

5, 6

6, 5

6, 6

$\small \textup{Probability}=\frac{3}{36}\:=\frac{1}{12}$

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Example Question #52 : Probability

10 cards, each with a distinct number from 1-10, are placed face-down on a table. If Matt chooses a card at random, what is the probability his card will display a prime number?

Possible Answers:

$\small \frac{2}{5}$

$\small \frac{3}{10}$

$\small \frac{1}{10}$

$\small \frac{5}{10}$

$\small \frac{1}{5}$

Correct answer:

$\small \frac{2}{5}$

Explanation:

Total outcomes: 10

Desired outcomes (prime numbers) = 4 (2, 3, 5, 7)

$\small \frac{4}{10}\:=\:\frac{2}{5}$

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PSAT Math : Statistics

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #42 : Probability

Example Question #44 : Probability

Example Question #43 : Probability

Example Question #163 : Data Analysis

Example Question #44 : Probability

Example Question #44 : Probability

Example Question #45 : Probability

Example Question #41 : Probability

Example Question #51 : Probability

Example Question #52 : Probability

All PSAT Math Resources

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