GMAT Math : Discrete Probability

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Example Question #1801 : Problem Solving Questions

Three coins are tossed at the same time. One of them is fair; two are loaded so that each comes up heads with probability . In terms of , what is the probability that the outcome will be three heads?

Possible Answers:

$\frac{3N}{8}$

$\frac{N ^{2} }{4}$

$\frac{N }{2}$

$\frac{N }{4}$

$\frac{N ^{2} }{2}$

Correct answer:

$\frac{N ^{2} }{2}$

Explanation:

The probability that all three coins will come up heads is the product of the individual probabilities:

$\frac{1}{2}\cdot N \cdot N = \frac{1}{2}N ^{2} = \frac{N ^{2} }{2}$

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Example Question #11 : Calculating Discrete Probability

One hundred marbles - each one red, yellow, blue, or green - are placed in a box. Forty of the marbles are green, there are twice as many blue marbles as there are red ones, and there are three times as many yellow marbles as red ones. What is the probability that a randomly drawn marble will be yellow?

Possible Answers:

$\frac{3}{20}$

$\frac{3}{10}$

$\frac{1}{10}$

$\frac{2}{5}$

$\frac{1}{5}$

Correct answer:

$\frac{3}{10}$

Explanation:

Let be the number of red marbles. Then there are blue marbles and yellow ones. Since there are 40 green marbles, and 100 marbles total, we can write this equation, simplifying and solving for :

x + 2x + 3x + 40 = 100

6x + 40 = 100

6x + 40 -40 = 100 -40

6x = 60

$6x \div 6 = 60\div 6$

x = 10

Therefore, there are $3x = 3 \cdot10 = 30$ yellow marbles, and the probability of drawing a yellow marble is $\frac{30}{100} = \frac{3}{10}$

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Example Question #1803 : Problem Solving Questions

Which of the following can be done to a standard deck of fifty-two playing cards without affecting the probability that a randomly chosen card will be a club?

Possible Answers:

Removing the hearts

Removing the ace of spades

Removing the fours

Removing the red kings

Adding the joker

Correct answer:

Removing the fours

Explanation:

The probability of drawing a club from a standard deck of cards is $\frac{1}{4}$ , since one-fourth of the cards - thirteen out of fifty-two - are clubs.

If the two red kings are removed, thirteen out of the remaining fifty cards are clubs, and the probability is $\frac{13}{50} \neq \frac{ 1}{4}$ .

If the thirteen hearts are removed, thirteen out of the remaining thirty-nine cards are clubs, and the probability is $\frac{13}{39}= \frac{ 1}{3} \neq \frac{ 1}{4}$ .

If the ace of spades is removed, thirteen out of the remaining fifty-one cards are clubs, and the probability is $\frac{13}{51} \neq \frac{ 1}{4}$ .

If the joker is added, thirteen out of the fifty-three cards are clubs, and the probability is $\frac{13}{53} \neq \frac{ 1}{4}$ .

If the four fours are removed, this leaves twelve clubs out of forty-eight cards, so the probability is $\frac{12}{48} = \frac{1}{4}$ . This is the corect choice.

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Example Question #1804 : Problem Solving Questions

Four coins are tossed at the same time. Three of them are fair; the fourth is loaded so that it comes up heads with probability . In terms of , what is the probability that the outcome will be three heads and one tail?

Possible Answers:

$\frac{1}{4} N + \frac{1}{8}$

$\frac{3}{8} N$

$\frac{3}{8} N + \frac{1}{8}$

$\frac{3}{8} N - \frac{1}{8}$

$\frac{1}{2} N - \frac{1}{8}$

Correct answer:

$\frac{1}{4} N + \frac{1}{8}$

Explanation:

Call the fair coins, which come up heads or tails with probability $\frac{1}{2}$ , Coins 1, 2, and 3; call the loaded coin, which comes up heads with probability and tails with probability 1-N , Coin 4. We are looking for the probability of one of the following four outcomes:

The probability of heads on Coins 1, 2, and 3 and tails on Coin 4:

$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot (1 -N) = \frac{1}{8} \cdot (1 -N) = \frac{1}{8} -\frac{1}{8} N$

The probability of tails on Coin 1 and heads on Coins 2, 3, and 4:

$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\cdot N= \frac{1}{8} N$

The probability of tails on Coin 2 and heads on Coins 1, 3, and 4:

$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\cdot N= \frac{1}{8} N$

The probability of tails on Coin 3 and heads on Coins 1, 2, and 4:

$\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\cdot N= \frac{1}{8} N$

Add these:

$\frac{1}{8} N +\frac{1}{8} N + \frac{1}{8} N + \frac{1}{8} -\frac{1}{8} N$

$= \frac{1}{8} N +\frac{1}{8} N + \frac{1}{8}$

$= \frac{1}{4} N + \frac{1}{8}$

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Example Question #11 : Calculating Discrete Probability

Two coins are loaded so that each comes up heads 60% of the time when tossed. If the coins are both tossed at the same time, what is the probability that the result will be one head and one tail?

Possible Answers:

0.6

0.48

0.24

0.52

0.5

Correct answer:

0.48

Explanation:

For each coin, the probability of a head is 0.6, and the probability of a tail is 0.4.

The probability of a head coming up on the first coin and a tail coming up on the second is $0.6 \cdot 0.4 = 0.24$ ; the same holds for the reverse case. Therefore, the probability of one head and one tail is

$\ 0.24 + 0.24 = 0.48$

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Example Question #12 : Calculating Discrete Probability

A die is loaded so that it is equally likely to come up 1, 2, 3, 4, or 5, but twice as likely to come up 6 as it is likely to come up 5. If it is rolled twice, what is the probability that both rolls are even?

Possible Answers:

$\frac{1}{4}$

$\frac{16}{49}$

$\frac{9} {49}$

$\frac{7}{18}$

$\frac{4}{7}$

Correct answer:

$\frac{16}{49}$

Explanation:

Let be the probability that the die will come up 1. Then is also the common probability for each of the rolls of 2, 3, 4, or 5, and is the probability of the die coming up 6. To determine the value of , add these six probabilities and set the sum to 1, then solve:

$5 \cdot P + 2P = 1$

7P = 1

$P = \frac{1}{7}$

So 1, 2, 3, 4, and 5 each will come up with probability $\frac{1}{7}$ and 6 will cme up with probability $\frac{2}{7}$ .

An even number will come up with probability $\frac{1}{7} + \frac{1}{7} + \frac{2}{7} = \frac{4}{7}$ . Two even rolls will happen with probability $\frac{4}{7 } \cdot \frac{4}{7 } = \frac{16}{49}$ .

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Example Question #13 : Calculating Discrete Probability

Fifty marbles are put into a big box: ten red, ten yellow, ten green, and twenty blue. Which of the following actions would change the probability that a random draw would result in a blue marble?

Possible Answers:

Removing two yellow marbles, four green marbles, and four blue marbles.

Removing half of the marbles of each color.

Removing three marbles of each color.

Adding three yellow marbles and two blue marbles.

Adding fifteen green marbles and ten blue marbles.

Correct answer:

Removing three marbles of each color.

Explanation:

The ratio of blue marbles to non-blue marbles is 20:30 = 2:3 . To maintain the same probability of drawing a blue marble, this ratio must remain the same after the action.

We will look at all five actions.

If three yellow marbles and two blue marbles are added, the ratio is 22:33 = 2:3 .

If fifteen green marbles and ten blue marbles are added, the ratio is 30:45 = 2:3 .

If two yellow marbles, four green marbles, and four blue marbles are removed, the ratio is 16:24 = 2:3 .

If half of the marbles of each color are removed - that is, ten blue marbles and five of each of the other three colors - the ratio is 10:15 = 2:3 .

If three marbles of each color are removed, however, the ratio is $17: 21\neq 2:3$ . This is the correct choice.

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Example Question #261 : Arithmetic

One hundred marbles - red, yellow, blue, or green - are placed in a box. There are an equal number of red and blue marbles, three times as many yellow marbles as green marbles, and five fewer green marbles than red marbles. What is the probability that a randomly drawn marble is NOT blue?

Possible Answers:

$\frac{4}{5}$

$\frac{11}{20}$

$\frac{3}{4}$

$\frac{17}{20}$

$\frac{3}{5}$

Correct answer:

$\frac{4}{5}$

Explanation:

If there are blue marbles, there are also red marbles, x-5 green marbles, and $3 \left (x-5 \right )$ yellow marbles. There are 100 marbles total, so we can solve for in the equation:

x + x + (x-5) + 3 (x-5) = 100

x + x + x-5 + 3 x-15= 100

6x -20 = 100

6x = 120

x = 20

20 of the 100 marbles are blue, so the probability that the marble is not blue is:

$\frac{100-20}{100} = \frac{80}{100} = \frac{4}{5}$

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Example Question #15 : 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 multiple of 3?

Possible Answers:

$\frac{5}{16}$

$\frac{25}{64}$

$\frac{23}{64}$

$\frac{11}{32}$

$\frac{3}{8}$

Correct answer:

$\frac{11}{32}$

Explanation:

There are $8 \cdot 8= 64$ possible outcomes. The outcome can be between 2 and 16 inclusive; we count the number of rolls that result in any of the possible multiples of 3 - 3,6,9,12,15:

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

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

9: (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1)

12: (4,8), (5,7), (6,6) , (7,5), (8.4)

15: (7,8), (8.7)

22 out of 64 outcomes result in a multiple of 3, so the probability is $\frac{22}{64} = \frac{11}{32}$

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Example Question #16 : 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 multiple of 5?

Possible Answers:

$\frac{3}{16}$

$\frac{15}{64}$

$\frac{13}{64}$

$\frac{7}{32}$

$\frac{11}{64}$

Correct answer:

$\frac{13}{64}$

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 any of the possible multiples of 5 - 5, 10, 15:

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

10: (2,8), (3,7), (4,6), (5,5) ,(6,4), (7,3), (8,2)

15: (7,8), (8.7)

13 out of 64 outcomes result in a multiple of 3, so the probability is $\frac{13}{64}$ .

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GMAT Math : Discrete Probability

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1801 : Problem Solving Questions

Example Question #11 : Calculating Discrete Probability

Example Question #1803 : Problem Solving Questions

Example Question #1804 : Problem Solving Questions

Example Question #11 : Calculating Discrete Probability

Example Question #12 : Calculating Discrete Probability

Example Question #13 : Calculating Discrete Probability

Example Question #261 : Arithmetic

Example Question #15 : Calculating Discrete Probability

Example Question #16 : Calculating Discrete Probability

All GMAT Math Resources

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