Discrete Probability

Description: Welcome to the Discrete Probability Quiz! Test your understanding of the fundamental concepts of probability in a discrete setting.
Number of Questions: 14
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Tags: discrete probability probability theory combinatorics
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In a standard deck of 52 cards, what is the probability of drawing an ace?

  1. (1/4)

  2. (1/13)

  3. (1/52)

  4. (4/13)


Correct Option: B
Explanation:

There are 4 aces in a standard deck of 52 cards. So, the probability of drawing an ace is 4/52 = 1/13.

A bag contains 10 red balls, 15 blue balls, and 5 green balls. If a ball is randomly drawn from the bag, what is the probability that it is not green?

  1. (25/30)

  2. (5/30)

  3. (10/30)

  4. (15/30)


Correct Option: A
Explanation:

The total number of balls in the bag is 10 + 15 + 5 = 30. The number of balls that are not green is 10 + 15 = 25. So, the probability of drawing a ball that is not green is 25/30.

A coin is tossed twice. What is the probability of getting two heads?

  1. (1/4)

  2. (1/2)

  3. (3/4)

  4. (1/8)


Correct Option: A
Explanation:

There are four possible outcomes when a coin is tossed twice: HH, HT, TH, TT. Only one of these outcomes, HH, results in two heads. So, the probability of getting two heads is 1/4.

A box contains 6 red balls, 4 blue balls, and 2 green balls. If two balls are randomly drawn from the box without replacement, what is the probability that both balls are red?

  1. (3/10)

  2. (1/5)

  3. (2/5)

  4. (1/2)


Correct Option: B
Explanation:

The probability of drawing a red ball on the first draw is 6/12 = 1/2. After the first draw, there are 5 red balls and 11 total balls remaining in the box. So, the probability of drawing a red ball on the second draw is 5/11. The probability of both events occurring is (1/2) * (5/11) = 1/5.

A survey of 100 people found that 60 people liked chocolate, 40 people liked vanilla, and 20 people liked both chocolate and vanilla. If a person is randomly selected from the survey, what is the probability that they like chocolate or vanilla?

  1. (80/100)

  2. (60/100)

  3. (40/100)

  4. (20/100)


Correct Option: A
Explanation:

Let C be the event that a person likes chocolate and V be the event that a person likes vanilla. Then, the probability that a person likes chocolate or vanilla is P(C or V) = P(C) + P(V) - P(C and V) = (60/100) + (40/100) - (20/100) = 80/100.

A company has 10 employees, 6 of whom are women and 4 of whom are men. If 3 employees are randomly selected for a project, what is the probability that all 3 are women?

  1. (20/120)

  2. (6/20)

  3. (3/20)

  4. (1/20)


Correct Option: D
Explanation:

The probability of selecting a woman on the first draw is 6/10. After the first draw, there are 5 women and 9 total employees remaining. So, the probability of selecting a woman on the second draw is 5/9. After the second draw, there are 4 women and 8 total employees remaining. So, the probability of selecting a woman on the third draw is 4/8. The probability of all three events occurring is (6/10) * (5/9) * (4/8) = 1/20.

A bag contains 10 white balls, 15 black balls, and 5 red balls. If 4 balls are randomly drawn from the bag without replacement, what is the probability that exactly 2 of the balls are white?

  1. (3/14)

  2. (1/7)

  3. (2/7)

  4. (4/7)


Correct Option: C
Explanation:

The probability of drawing a white ball on the first draw is 10/30. After the first draw, there are 9 white balls and 29 total balls remaining in the bag. So, the probability of drawing a white ball on the second draw is 9/29. After the second draw, there are 8 white balls and 28 total balls remaining in the bag. So, the probability of drawing a white ball on the third draw is 8/28. After the third draw, there are 7 white balls and 27 total balls remaining in the bag. So, the probability of drawing a white ball on the fourth draw is 7/27. The probability of exactly 2 of the balls being white is (10/30) * (9/29) * (8/28) * (19/27) = 2/7.

A fair six-sided die is rolled twice. What is the probability of getting a sum of 7?

  1. (1/6)

  2. (1/3)

  3. (1/2)

  4. (5/12)


Correct Option: D
Explanation:

There are 36 possible outcomes when a fair six-sided die is rolled twice. The outcomes that result in a sum of 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). So, the probability of getting a sum of 7 is 6/36 = 1/6.

A company has 12 employees, 8 of whom are women and 4 of whom are men. If 5 employees are randomly selected for a meeting, what is the probability that at least one of the employees is a woman?

  1. (1/2)

  2. (2/5)

  3. (3/5)

  4. (4/5)


Correct Option: D
Explanation:

The probability that none of the employees selected are women is (4/12) * (3/11) * (2/10) * (1/9) * (0/8) = 0. The probability that at least one of the employees selected is a woman is 1 - 0 = 1.

A bag contains 10 red balls, 15 blue balls, and 5 green balls. If 3 balls are randomly drawn from the bag without replacement, what is the probability that all 3 balls are of different colors?

  1. (1/14)

  2. (2/7)

  3. (3/14)

  4. (4/7)


Correct Option: C
Explanation:

The probability of drawing a red ball on the first draw is 10/30. After the first draw, there are 9 red balls, 15 blue balls, and 4 green balls remaining in the bag. So, the probability of drawing a blue ball on the second draw is 15/29. After the second draw, there are 9 red balls, 14 blue balls, and 4 green balls remaining in the bag. So, the probability of drawing a green ball on the third draw is 4/28. The probability of all three events occurring is (10/30) * (15/29) * (4/28) = 3/14.

A fair coin is tossed 10 times. What is the probability of getting exactly 5 heads?

  1. (1/1024)

  2. (1/512)

  3. (1/256)

  4. (1/128)


Correct Option: C
Explanation:

The probability of getting a head on a single toss is 1/2. The probability of getting a tail on a single toss is also 1/2. The probability of getting exactly 5 heads in 10 tosses is given by the binomial distribution: P(X = 5) = (10 choose 5) * (1/2)^5 * (1/2)^5 = 252 * 1/32 = 1/256.

A box contains 12 red balls, 18 blue balls, and 6 green balls. If 4 balls are randomly drawn from the box without replacement, what is the probability that there are exactly 2 red balls and 2 blue balls?

  1. (1/14)

  2. (1/7)

  3. (2/7)

  4. (3/14)


Correct Option: C
Explanation:

The probability of drawing a red ball on the first draw is 12/36. After the first draw, there are 11 red balls, 18 blue balls, and 6 green balls remaining in the box. So, the probability of drawing a red ball on the second draw is 11/35. After the second draw, there are 10 red balls, 18 blue balls, and 6 green balls remaining in the box. So, the probability of drawing a blue ball on the third draw is 18/34. After the third draw, there are 10 red balls, 17 blue balls, and 6 green balls remaining in the box. So, the probability of drawing a blue ball on the fourth draw is 17/33. The probability of all four events occurring is (12/36) * (11/35) * (18/34) * (17/33) = 2/7.

A company has 20 employees, 12 of whom are women and 8 of whom are men. If 5 employees are randomly selected for a project, what is the probability that the project team consists of exactly 3 women and 2 men?

  1. (1/10)

  2. (1/5)

  3. (2/5)

  4. (3/10)


Correct Option: D
Explanation:

The probability of selecting 3 women on the first, second, and third draws is (12/20) * (11/19) * (10/18) = 110/342. The probability of selecting 2 men on the fourth and fifth draws is (8/17) * (7/16) = 56/272. The probability of all five events occurring is (110/342) * (56/272) = 3/10.

A bag contains 15 red balls, 10 blue balls, and 5 green balls. If 4 balls are randomly drawn from the bag without replacement, what is the probability that there are more red balls than blue balls?

  1. (1/2)

  2. (2/5)

  3. (3/5)

  4. (4/5)


Correct Option: C
Explanation:

There are 5 possible outcomes where there are more red balls than blue balls: (3 red, 1 blue), (3 red, 1 green), (2 red, 2 blue), (2 red, 1 green, 1 blue), and (1 red, 3 blue). The probability of each of these outcomes is given by the multinomial distribution: P(X = 3 red, 1 blue) = (4 choose 3) * (15 choose 3) * (10 choose 1) / (30 choose 4) = 1350 / 27405, P(X = 3 red, 1 green) = (4 choose 3) * (15 choose 3) * (5 choose 1) / (30 choose 4) = 675 / 27405, P(X = 2 red, 2 blue) = (4 choose 2) * (15 choose 2) * (10 choose 2) / (30 choose 4) = 2700 / 27405, P(X = 2 red, 1 green, 1 blue) = (4 choose 2) * (15 choose 2) * (5 choose 1) * (10 choose 1) / (30 choose 4) = 1350 / 27405, and P(X = 1 red, 3 blue) = (4 choose 1) * (15 choose 1) * (10 choose 3) / (30 choose 4) = 675 / 27405. The probability of more red balls than blue balls is the sum of these probabilities: 1350/27405 + 675/27405 + 2700/27405 + 1350/27405 + 675/27405 = 6750/27405 = 3/5.

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