GMAT Math : Discrete Probability

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Example Question #21 : Calculating Discrete Probability

Target_1

The above figure depicts a square target. It is divided into sixteen squares of equal area, one of which is in turn divided into two congruent triangles.

According to the rules of a game, if a dart is thrown at this target, the number of points scored for each color are as follows:

Gray: 1 point

Red: 2 points

Yellow: 4 points

Green: 7 points

Blue: 10 points.

Two darts are thrown at random at the target. Assuming that both darts hit the board, and assuming that there is no skill involved, what is the probability that more than 15 points will be scored?

Possible Answers:

$\frac{5}{512}$

$\frac{1}{ 256}$

$\frac{1}{ 128}$

$\frac{1}{16}$

$\frac{7}{ 1,024}$

Correct answer:

$\frac{7}{ 1,024}$

Explanation:

One of two things must happen for a score of 15 or better to be made: either both darts must land on blue (20 points), or one dart must land on green and one on blue, in either order (17 points). The next-highest possiblities are two green (14 points) or one yellow and one blue (14 points).

Since one half of a square out of sixteen is blue, the probability that a randomly thrown dart will hit blue is $\frac{\frac{1}{2}}{16} = \frac{1}{32}$ .

Since one and one-half squares out of sixteen are green, the probability that a randomly thrown dart will hit green is $\frac{\frac{3}{2}}{16} = \frac{3}{32}$ .

The probability that both darts will hit blue is $\frac{1}{32} \cdot \frac{1}{32} = \frac{1}{ 1,024}$ .

The probability that the first dart will hit blue and the second will hit green is $\frac{1}{32} \cdot \frac{3}{32} = \frac{3}{ 1,024}$ , which is also the probability that the reverse will happen.

Add these probabilities:

$\frac{1}{ 1,024} + \frac{3}{ 1,024}+ \frac{3}{ 1,024} = \frac{7}{ 1,024}$

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Example Question #22 : Calculating Discrete Probability

Two eight-sided dice from a role-playing game are thrown. Each die is fair and marked with the numbers 1 through 8. What is the probability that the sum of the dice will be a prime number?

Possible Answers:

$\frac{21}{64}$

$\frac{23}{64}$

$\frac{19}{64}$

$\frac{25}{64}$

$\frac{27}{64}$

Correct answer:

$\frac{23}{64}$

Explanation:

There are $8 \cdot 8= 64$ possible outcomes. The sum can be between 2 and 16 inclusive; we count the number of rolls that result in the sum being any of the possible prime numbers 2,3,5,7,11,13:

2: (1,1)

3: (1,2), (2,1)

5: (1,4), (2,3), (3,2), (4,1)

7: (1,6), (2,5), (3,4) , (4,3), (5,2), (6,1)

11: (3,8), (4,7), (5,6), (6,5), (7,4),(8,3)

13 : (5,8), (6,7), (7,6), (8,5)

23 out of 64 rolls result in prime sums, so the probability is $\frac{23}{64}$ .

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Example Question #23 : Calculating Discrete Probability

A box contains 12 balls, of which 5 are black, 4 are red, and 3 are white. If 2 balls are randomly selected from the box, one at a time without being replaced, what is the probability that the first ball selected will be red and the second ball selected will be white?

Possible Answers:

$\frac{1}{36}$

$\frac{1}{11}$

$\frac{1}{7}$

$\frac{1}{12}$

$\frac{7}{12}$

Correct answer:

$\frac{1}{11}$

Explanation:

The probability that the first ball selected will be red is $\frac{4}{12}=\frac{1}{3}$ .

The probability that the second ball selected will be white is $\frac{3}{11}$ .

We can solve by the multiplication principle since the two events happen together.

$\left ( \frac{1}{3} \right )\times\left ( \frac{3}{11} \right )=\frac{1}{11}$

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Example Question #21 : Calculating Discrete Probability

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

Possible Answers:

$\frac{7}{32}$

$\frac{1}{4}$

$\frac{7}{64}$

$\frac{5}{18}$

$\frac{1}{8}$

Correct answer:

$\frac{7}{32}$

Explanation:

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \},\left \{ 2,3\right \}, \left \{ 3,4\right \} ,\left \{ 4,5\right \}, \left \{ 5,6\right \} , \left \{ 6,7\right \}, \left \{ 7,8\right \},$

$\left \{ 2,1\right \},\left \{ 3,2\right \}, \left \{ 4,3\right \} ,\left \{ 5,4\right \}, \left \{ 6,5\right \} , \left \{ 7,6\right \}, \left \{ 8,7\right \}$

This makes 14 favorable outcomes out of 64 outcomes total, so the probability is

$\frac{14}{64} = \frac{7}{32}$ .

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Example Question #25 : Calculating Discrete Probability

Three boxes contain marbles, each one either red or white.

Box 1 contains 20 red marbles and 10 white marbles.

Box 2 contains 30 red marbles and 10 white marbles.

Box 3 conatins 40 red marbles and 10 white marbles.

One of three boxes is selected at random, and one marble is selected from that box at random. What is the probability that a white marble will be selected?

Possible Answers:

$\frac{47}{90}$

$\frac{1}{4}$

$\frac{47}{180}$

$\frac{1}{3}$

$\frac{47}{120}$

Correct answer:

$\frac{47}{180}$

Explanation:

This is a conditional probability problem.

Each box can be selected with probability $\frac{1}{3}$ .

If Box 1 is selected, the probability of selecting a white marble is $\frac{10}{30} =\frac{1}{3}$ . The overall probability of selecting Box 1, then a white marble, is $\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$ .

If Box 2 is selected, the probability of selecting a white marble is $\frac{10}{40} =\frac{1}{4}$ . The overall probability of selecting Box 2, then a white marble, is $\frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$ .

If Box 3 is selected, the probability of selecting a white marble is $\frac{10}{50} =\frac{1}{5}$ . The overall probability of selecting Box 2, then a white marble, is $\frac{1}{3} \cdot \frac{1}{5} = \frac{1}{15}$ .

The overall probability of selecting a white marble is the sum of these probabilities:

$\frac{1}{9} +\frac{1}{12} + \frac{1}{15} = \frac{20}{180} +\frac{15}{180} + \frac{12}{180}= \frac{47}{180}$

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Example Question #26 : Calculating Discrete Probability

Two fair six-sided dice are altered. One of them has its "6" changed to a "1"; the other has its "1" changed to a "6". The dice are tossed and their sum is noted.

What is the probability that the sum will be 7?

Possible Answers:

$\frac{7}{36}$

$\frac{2}{9}$

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{6}$

Correct answer:

$\frac{2}{9}$

Explanation:

We will call the faces of the first die (the one with two 1's and no 6) 1A, 1B, 2, 3, 4, 5; we will call the faces of the second die (the one with two 6's and no 1) 2, 3, 4, 5, 6A, 6B.

A seven can be rolled in the following ways - with the outcome of the first die and the second die listed in that order:

$\left \{ 1A, 6A\right \}$

$\left \{ 1A, 6B\right \}$

$\left \{ 1B, 6A\right \}$

$\left \{ 1B, 6B\right \}$

$\left \{ 2,5\right \}$

$\left \{ 3,4\right \}$

$\left \{ 4,3\right \}$

$\left \{ 5,2\right \}$

Note that a 6 cannot be rolled on the first die, nor can a 1 be rolled on the second.

This makes 8 rolls out of 36 that can result in a 7; since the dice are fair, the probability of rolling one of these results is $\frac{8}{36} = \frac{2}{9}$

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Example Question #21 : Discrete Probability

What is the probability of drawing a red king in a standard card deck?

Possible Answers:

$\frac{1}{26}$

$\frac{4}{13}$

$\frac{1}{52}$

$\frac{1}{13}$

Correct answer:

$\frac{1}{26}$

Explanation:

There are red kings in a standard deck of cards, therefore the probability of drawing red kings is $\frac{2}{52}$ which simplifies to $\frac{1}{26}$ .

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Example Question #28 : Calculating Discrete Probability

A box contains tickets numbered 01-50 . What is the probability of randomly selecting a ticket that has a in the ones place?

Possible Answers:

$50\%$

$5\%$

$10\%$

$20\%$

Correct answer:

$10\%$

Explanation:

From 01-50 , there are tickets that will have a in the ones place: 2,12,22,32,42 .

Therefore the probability of drawing a ticket with a in the ones place is $\frac{5}{50}$ which is 10%.

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Example Question #21 : Calculating Discrete Probability

If N is a number chosen at random from the set [3, -7, 12, 8, 11] , and P is a number chosen at random from the set [1, 2, 0, 9] , what is the probability that

N+P=12 ?

Possible Answers:

0.15

0.85

0.25

0.12

0.24

Correct answer:

0.15

Explanation:

The total possible pairs of numbers for the two sets is 20, since there are 5 numbers in the first set and 4 numbers in the second set. Total possible outcomes are found by multiplying the number of terms in each set together.

5x4=20, so there are 20 possible pairs.

How many pairs sum to 12?

There are three pairs that work: 3 from the first set and 9 from the second, 12 from the first set and 0 from the second, and 11 from the first set and 1 from the second.

Since 3 out of 20 pairs sum to twelve, the probability of N+P=12 is:

$\frac{3}{20}=0.15$

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Example Question #30 : Calculating Discrete Probability

Jane has toy cars. of them are trucks and are race cars. are blue, are green, are orange, and are red.

What is the probability of having a blue car OR a race car?

Possible Answers:

$\frac{17}{20}$

$\frac{3}{20}$

$\frac{7}{10}$

$\frac{12}{20}$

$\frac{7}{20}$

Correct answer:

$\frac{7}{10}$

Explanation:

P(Blue Car OR Race Car) = P(Blue Car) + P(Race Car) - P(Blue Race Car)

P(Blue Car) = 5/20

P(Race Car) = 12/20

P(Blue Race Car) $= 5/20 \times 12/20 = 1/4 \times 3/5 = 3/20$

So P(Blue Car OR Race Car) = 5/20 + 12/20 - 3/20 = (5+12-3)/20 = 14/20 = 7/10

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GMAT Math : Discrete Probability

Study concepts, example questions & explanations for GMAT Math

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Example Questions

Example Question #21 : Calculating Discrete Probability

Example Question #22 : Calculating Discrete Probability

Example Question #23 : Calculating Discrete Probability

Example Question #21 : Calculating Discrete Probability

Example Question #25 : Calculating Discrete Probability

Example Question #26 : Calculating Discrete Probability

Example Question #21 : Discrete Probability

Example Question #28 : Calculating Discrete Probability

Example Question #21 : Calculating Discrete Probability

Example Question #30 : Calculating Discrete Probability

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