Probability with Replacement
Probability with replacement is a type of probability where an object is selected from a set, and then returned back to the set before the next selection is made. This means that the probability of each selection is not affected by previous selections. The formula for probability with replacement is:
P(E) = n(E) / n(S)
where P(E) is the probability of an event E occurring, n(E) is the number of ways that event E can occur, and n(S) is the total number of outcomes in the sample space.
Probability without Replacement
Probability without replacement is a type of probability where an object is selected from a set, and then not returned back to the set before the next selection is made. This means that the probability of each selection is affected by previous selections. The formula for probability without replacement is:
P(E) = (n(E) / n(S)) x [(n(E) – 1) / (n(S) – 1)] x … x [(n(E) – k + 1) / (n(S) – k + 1)]
where P(E) is the probability of an event E occurring, n(E) is the number of ways that event E can occur, n(S) is the total number of outcomes in the sample space, and k is the number of selections made. The notation “n(E) – k + 1” means that after each selection, the number of objects in the set decreases by 1, so we need to adjust the number of ways that event E can occur accordingly.
Practice Problems on Probability with Replacement
A fair six-sided die is rolled twice. What is the probability of getting a 3 on both rolls?
Solution:
The probability of rolling a 3 on one roll is 1/6.
Since the rolls are independent and with replacement, the probability of rolling a 3 on both rolls is (1/6) x (1/6) = 1/36.
A jar contains 5 red balls and 3 blue balls. A ball is randomly selected from the jar, its color is noted, and then it is returned to the jar. This process is repeated twice. What is the probability that both balls are blue?
Solution:
The probability of selecting a blue ball on one draw is 3/8.
Since the draws are independent and with replacement, the probability of selecting two blue balls is (3/8) x (3/8) = 9/64.
A factory produces light bulbs, and 4% of the bulbs are defective. A sample of 20 bulbs is randomly selected with replacement. What is the probability that at least one bulb is defective?
Solution:
The probability of selecting a non-defective bulb on one draw is 0.96.
Since the draws are independent and with replacement, the probability of selecting only non-defective bulbs is 0.96^20 ≈ 0.32.
Therefore, the probability of selecting at least one defective bulb is 1 – 0.32 = 0.68.
A bag contains 7 red marbles, 4 blue marbles, and 2 green marbles. A marble is randomly selected from the bag, its color is noted, and then it is returned to the bag. This process is repeated three times. What is the probability of getting one red marble, one blue marble, and one green marble, in any order?
Solution:
The probability of selecting a red marble on one draw is 7/13.
The probability of selecting a blue marble on one draw is 4/13.
The probability of selecting a green marble on one draw is 2/13.
Since the draws are independent and with replacement, the probability of getting one red, one blue, and one green marble, in any order, is (7/13) x (4/13) x (2/13) x 3! = 0.0915.
A deck of 52 playing cards is shuffled and a card is drawn. The card is then returned to the deck and the deck is shuffled again. What is the probability of drawing a red card both times?
Solution:
The probability of drawing a red card on one draw is 26/52 = 1/2.
Since the draws are independent and with replacement, the probability of drawing a red card both times is (1/2) x (1/2) = 1/4.
A box contains 12 red balls, 8 blue balls, and 5 green balls. A ball is randomly selected from the box, its color is noted, and then it is returned to the box. This process is repeated four times. What is the probability of getting two red balls, one blue ball, and one green ball, in any order?
Solution:
The probability of selecting a red ball on one draw is 12/25.
The probability of selecting a blue ball on one draw is 8/25.
The probability of selecting a green ball on one draw is 5/25 = 1/5.
Since the draws are independent and with replacement, the probability of getting two red balls, one blue ball, one green ball, in any order, is:
(12/25)^2 x (8/25) x (1/5) x 4! = 0.04224
A box contains 10 black balls and 5 white balls. A ball is randomly selected from the box, its color is noted, and then it is returned to the box. This process is repeated three times. What is the probability of getting at least one white ball?
Solution:
The probability of selecting a white ball on one draw is 5/15 = 1/3.
Since the draws are independent and with replacement, the probability of not selecting a white ball in three draws is (2/3)^3 = 8/27.
Therefore, the probability of selecting at least one white ball is 1 – 8/27 = 19/27.
A jar contains 6 red candies, 5 blue candies, and 4 green candies. A candy is randomly selected from the jar, its color is noted, and then it is returned to the jar. This process is repeated five times. What is the probability of getting exactly 2 red candies and 3 blue candies, in any order?
Solution:
The probability of selecting a red candy on one draw is 6/15 = 2/5.
The probability of selecting a blue candy on one draw is 5/15 = 1/3.
Since the draws are independent and with replacement, the probability of getting exactly 2 red candies and 3 blue candies, in any order, is:
(2/5)^2 x (1/3)^3 x 5! / (2! x 3!) = 0.1248
A jar contains 3 black balls and 4 white balls. A ball is randomly selected from the jar, its color is noted, and then it is returned to the jar. This process is repeated four times. What is the probability of getting two black balls and two white balls, in any order?
Solution:
The probability of selecting a black ball on one draw is 3/7.
The probability of selecting a white ball on one draw is 4/7.
Since the draws are independent and with replacement, the probability of getting two black balls and two white balls, in any order, is:
(3/7)^2 x (4/7)^2 x 4! / (2! x 2!) = 0.297
A jar contains 10 red marbles, 8 blue marbles, and 6 green marbles. A marble is randomly selected, and then replaced. What is the probability that two marbles randomly selected in succession are both blue?
Solution:
The probability of selecting a blue marble on the first draw is 8/24.
The probability of selecting another blue marble on the second draw, with replacement, is also 8/24.
Therefore, the probability of selecting two blue marbles in succession is (8/24) x (8/24) = 16/72.
A bag contains 5 black balls and 7 white balls. A ball is randomly selected, and then replaced. What is the probability that three balls randomly selected in succession are all black?
Solution:
The probability of selecting a black ball on the first draw is 5/12.
The probability of selecting another black ball on the second draw, with replacement, is also 5/12.
The probability of selecting a third black ball on the third draw, with replacement, is also 5/12.
Therefore, the probability of selecting three black balls in succession is (5/12) x (5/12) x (5/12) = 125/1728.
A box contains 2 red balls and 3 blue balls. A ball is randomly selected, and then replaced. What is the probability that the first ball is red and the second ball is blue?
Solution:
The probability of selecting a red ball on the first draw is 2/5.
The probability of selecting a blue ball on the second draw, with replacement, is 3/5.
Therefore, the probability of selecting a red ball first and a blue ball second is (2/5) x (3/5) = 6/25.
A box contains 4 white balls and 6 black balls. A ball is randomly selected, and then replaced. What is the probability that the first ball is white and the second ball is also white?
Solution:
The probability of selecting a white ball on the first draw is 4/10.
The probability of selecting another white ball on the second draw, with replacement, is also 4/10.
Therefore, the probability of selecting two white balls in succession is (4/10) x (4/10) = 16/100.
A bag contains 8 red candies and 5 green candies. A candy is randomly selected, and then replaced. What is the probability that four candies randomly selected in succession are all green?
Solution:
The probability of selecting a green candy on the first draw is 5/13.
The probability of selecting another green candy on the second draw, with replacement, is also 5/13.
The probability of selecting a third green candy on the third draw, with replacement, is also 5/13.
The probability of selecting a fourth green candy on the fourth draw, with replacement, is also 5/13.
Therefore, the probability of selecting four green candies in succession is (5/13) x (5/13) x (5/13) x (5/13) = 625/28561.
A jar contains 12 red marbles and 8 blue marbles. A marble is randomly selected, and then replaced. What is the probability that five marbles randomly selected in succession are all red?
Solution:
The probability of selecting a red marble on the first draw is 12/20.
The probability of selecting another red marble on the second draw, with replacement, is also 12/20.
The probability of selecting a third red marble on the third draw, with replacement, is also 12/20.
The probability of selecting a fourth red marble on the fourth draw, with replacement, is also 12/20.
The probability of selecting a fifth red marble on the fifth draw, with replacement, is also 12/20.
Therefore, the probability of selecting five red marbles in succession is (12/20) x (12/20) x (12/20) x (12/20) x (12/20) = 248832/320000 = 39/50.
Practice Problems on Probability without Replacement
A bag contains 3 red marbles and 5 blue marbles. If two marbles are randomly selected without replacement, what is the probability that both marbles are red?
Solution:
The probability of selecting a red marble on the first draw is 3/8.
The probability of selecting a red marble on the second draw, without replacement, is 2/7.
Therefore, the probability of selecting two red marbles is (3/8) x (2/7) = 3/28.
A box contains 4 red balls, 3 blue balls, and 2 green balls. Two balls are randomly selected without replacement. What is the probability that one ball is red and one ball is blue?
Solution:
The probability of selecting a red ball on the first draw is 4/9.
The probability of selecting a blue ball on the second draw, without replacement, is 3/8.
The probability of selecting a blue ball on the first draw is 3/9.
The probability of selecting a red ball on the second draw, without replacement, is 4/8.
Therefore, the probability of selecting one red ball and one blue ball is (4/9) x (3/8) + (3/9) x (4/8) = 2/3.
A group of 6 people consists of 3 men and 3 women. If 3 people are randomly selected without replacement, what is the probability that all 3 are women?
Solution:
The probability of selecting a woman on the first draw is 3/6.
The probability of selecting a woman on the second draw, without replacement, is 2/5.
The probability of selecting a woman on the third draw, without replacement, is 1/4.
Therefore, the probability of selecting all 3 women is (3/6) x (2/5) x (1/4) = 1/20.
A box contains 10 red marbles, 6 blue marbles, and 4 green marbles. Three marbles are randomly selected without replacement. What is the probability that all 3 marbles are green?
Solution:
The probability of selecting a green marble on the first draw is 4/20.
The probability of selecting a green marble on the second draw, without replacement, is 3/19.
The probability of selecting a green marble on the third draw, without replacement, is 2/18.
Therefore, the probability of selecting all 3 green marbles is (4/20) x (3/19) x (2/18) = 1/285.
A deck of 52 playing cards is shuffled and two cards are randomly selected without replacement. What is the probability that both cards are aces?
Solution:
The probability of selecting an ace on the first draw is 4/52.
The probability of selecting an ace on the second draw, without replacement, is 3/51.
Therefore, the probability of selecting two aces is (4/52) x (3/51) = 1/221.
A bag contains 7 red marbles and 5 blue marbles. If two marbles are randomly selected without replacement, what is the probability that both marbles are red?
Solution:
The probability of selecting a red marble on the first draw is 7/12.
The probability of selecting a red marble on the second draw, without replacement, is 6/11.
Therefore, the probability of selecting two red marbles is (7/12) x (6/11) = 7/22.
A box contains 5 black balls and 3 white balls. Three balls are randomly selected without replacement. What is the probability that all 3 balls are black?
Solution:
The probability of selecting a black ball on the first draw is 5/8.
The probability of selecting a black ball on the second draw, without replacement, is 4/7.
The probability of selecting a black ball on the third draw, without replacement, is 3/6.
Therefore, the probability of selecting all 3 black balls is (5/8) x (4/7) x (3/6) = 5/28.
A jar contains 8 red marbles and 6 green marbles. Two marbles are randomly selected without replacement. What is the probability that both marbles are red?
Solution:
The probability of selecting a red marble on the first draw is 8/14.
The probability of selecting a red marble on the second draw, without replacement, is 7/13.
Therefore, the probability of selecting two red marbles is (8/14) x (7/13) = 28/91.
A group of 10 students consists of 5 boys and 5 girls. If 2 students are randomly selected without replacement, what is the probability that both students are girls?
Solution:
The probability of selecting a girl on the first draw is 5/10.
The probability of selecting a girl on the second draw, without replacement, is 4/9.
Therefore, the probability of selecting two girls is (5/10) x (4/9) = 2/9.
A box contains 3 red balls, 2 blue balls, and 1 green ball. Two balls are randomly selected without replacement. What is the probability that both balls are blue?
Solution:
The probability of selecting a blue ball on the first draw is 2/6.
The probability of selecting a blue ball on the second draw, without replacement, is 1/5.
Therefore, the probability of selecting two blue balls is (2/6) x (1/5) = 1/15.
A box contains 6 red balls and 4 green balls. Two balls are randomly selected without replacement. What is the probability that both balls are green?
Solution:
The probability of selecting a green ball on the first draw is 4/10.
The probability of selecting another green ball on the second draw, without replacement, is 3/9.
Therefore, the probability of selecting two green balls is (4/10) x (3/9) = 2/15.
A bag contains 7 blue marbles, 4 red marbles, and 3 green marbles. Three marbles are randomly selected without replacement. What is the probability that all 3 marbles are blue?
Solution:
The probability of selecting a blue marble on the first draw is 7/14.
The probability of selecting another blue marble on the second draw, without replacement, is 6/13.
The probability of selecting a third blue marble on the third draw, without replacement, is 5/12.
Therefore, the probability of selecting three blue marbles is (7/14) x (6/13) x (5/12) = 35/156.
A group of 15 people consists of 8 men and 7 women. If 3 people are randomly selected without replacement, what is the probability that all 3 people are men?
Solution:
The probability of selecting a man on the first draw is 8/15.
The probability of selecting another man on the second draw, without replacement, is 7/14.
The probability of selecting a third man on the third draw, without replacement, is 6/13.
Therefore, the probability of selecting three men is (8/15) x (7/14) x (6/13) = 8/65.
A box contains 4 white balls, 3 red balls, and 2 blue balls. Two balls are randomly selected without replacement. What is the probability that the first ball is red and the second ball is blue?
Solution:
The probability of selecting a red ball on the first draw is 3/9.
The probability of selecting a blue ball on the second draw, without replacement, is 2/8.
Therefore, the probability of selecting a red ball first and a blue ball second is (3/9) x (2/8) = 1/12.
A deck of cards contains 52 cards, including 4 aces. Two cards are randomly selected without replacement. What is the probability that both cards are aces?
Solution:
The probability of selecting an ace on the first draw is 4/52.
The probability of selecting another ace on the second draw, without replacement, is 3/51.
Therefore, the probability of selecting two aces is (4/52) x (3/51) = 1/221.
General Probability
A box contains 10 red balls and 5 blue balls. If two balls are drawn at random, what is the probability that both are red?
Solution:
Total number of balls in the box = 10 + 5 = 15
Probability of selecting a red ball on the first draw = 10/15
Probability of selecting a red ball on the second draw = 9/14 (since one red ball has already been selected)
Therefore, the probability of selecting two red balls = (10/15) x (9/14) = 3/7
In a class of 30 students, 10 are girls and 20 are boys. If a student is selected at random, what is the probability that the student is a girl?
Solution:
Total number of students in the class = 30
Probability of selecting a girl = 10/30 = 1/3
A bag contains 4 red balls and 6 blue balls. If two balls are drawn at random without replacement, what is the probability that one ball is red and the other is blue?
Solution:
Total number of balls in the bag = 4 + 6 = 10
Probability of selecting a red ball on the first draw = 4/10
Probability of selecting a blue ball on the second draw = 6/9 (since one red ball has already been selected)
Probability of selecting a blue ball on the first draw = 6/10
Probability of selecting a red ball on the second draw = 4/9 (since one blue ball has already been selected)
Therefore, the probability of selecting one red and one blue ball = (4/10 x 6/9) + (6/10 x 4/9) = 24/90 = 4/15
A bag contains 5 red balls and 3 blue balls. If two balls are drawn at random with replacement, what is the probability that both balls are red?
Solution:
Total number of balls in the bag = 5 + 3 = 8
Probability of selecting a red ball on the first draw = 5/8
Probability of selecting a red ball on the second draw = 5/8 (since the first ball is replaced)
Therefore, the probability of selecting two red balls = (5/8) x (5/8) = 25/64
A fair die is rolled twice. What is the probability that the sum of the two rolls is 7?
Solution:
Total number of outcomes = 6 x 6 = 36
Number of outcomes where the sum is 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} = 6
Therefore, the probability of getting a sum of 7 = 6/36 = 1/6
A coin is flipped 4 times. What is the probability of getting exactly 3 heads?
Solution:
Total number of outcomes = 2 x 2 x 2 x 2 = 16
Number of outcomes where exactly 3 heads are obtained = 4
Therefore, the probability of getting exactly 3 heads = 4/16 = 1/4
A jar contains 5 red balls and 3 green balls. If 2 balls are drawn at random without replacement, what is the probability that both balls are green?
Solution:
Total number of balls in the jar = 5 + 3 = 8
Probability of selecting a green ball on the first draw = 3/8
Probability of selecting a green ball on the second draw = 2/7 (since one green ball has already been selected)
Therefore, the probability of selecting two green balls = (3/8) x (2/7) = 3/28
A bag contains 7 black balls, 5 white balls, and 4 red balls. If two balls are drawn at random without replacement, what is the probability that one ball is black and one ball is white?
Solution:
Total number of balls in the bag = 7 + 5 + 4 = 16
Number of ways to select one black ball and one white ball = 7 x 5 = 35
Number of ways to select two balls from the bag = 16 x 15 = 240
Therefore, the probability of selecting one black ball and one white ball = 35/240 = 7/48
A box contains 12 balls, out of which 3 are red, 4 are blue, and 5 are green. If 3 balls are drawn at random without replacement, what is the probability that all 3 are green?
Solution:
Total number of balls in the box = 12
Number of ways to select 3 green balls = 5C3 = 10
Number of ways to select 3 balls from the box = 12C3 = 220
Therefore, the probability of selecting all 3 green balls = 10/220 = 1/22
A factory produces bulbs, and 5% of the bulbs are defective. If 3 bulbs are selected at random, what is the probability that all 3 are non-defective?
Solution:
Probability of selecting a non-defective bulb = 1 – 0.05 = 0.95
Probability of selecting 3 non-defective bulbs = (0.95)^3 = 0.857375
In a deck of 52 playing cards, what is the probability of drawing a queen or a king?
Solution:
Number of queens and kings in the deck = 4 + 4 = 8
Total number of cards in the deck = 52
Probability of drawing a queen or a king = 8/52 = 2/13
A bag contains 3 black balls and 2 white balls. If 2 balls are drawn at random with replacement, what is the probability that both balls are black?
Solution:
Total number of balls in the bag = 3 + 2 = 5
Probability of selecting a black ball on the first draw = 3/5
Probability of selecting a black ball on the second draw = 3/5 (since the first ball is replaced)
Therefore, the probability of selecting two black balls = (3/5) x (3/5) = 9/25
In a group of 20 people, what is the probability that at least 2 people have the same birthday?
Solution:
Number of possible birthdays = 365
Probability that no two people have the same birthday = 365/365 x 364/365 x 363/365 x … x 346/365 (since the first person can have any birthday, the second person can have any of the remaining 364, the third person can have any of the remaining 363, and so on)
Therefore, the probability that at least 2 people have the same birthday = 1 – (365/365 x 364/365 x 363/365 x … x 346/365)
In a bag, there are 8 red balls, 6 green balls, and 4 blue balls. If a ball is drawn at random, what is the probability that it is not blue?
Solution:
Total number of balls in the bag = 8 + 6 + 4 = 18
Number of non-blue balls = 8 + 6 = 14
Therefore, the probability of selecting a non-blue ball = 14/18 = 7/9
A coin is tossed 4 times. What is the probability of getting at least one head?
Solution:
Total number of possible outcomes = 2^4 = 16
Number of outcomes in which all 4 tosses are tails = 1
Therefore, the probability of getting at least one head = 1 – (1/16) = 15/16
A company produces laptops, and 5% of the laptops are defective. A retailer purchases 10 laptops from the company. What is the probability that at least one laptop is defective?
Solution:
Probability of a laptop being non-defective = 1 – 0.05 = 0.95
Probability that all 10 laptops are non-defective = (0.95)^10 = 0.5987369392
Therefore, the probability that at least one laptop is defective = 1 – 0.5987369392 = 0.4012630608
In a bag, there are 4 red balls, 3 green balls, and 5 blue balls. If 3 balls are drawn at random without replacement, what is the probability that all 3 balls are blue?
Solution:
Total number of balls in the bag = 4 + 3 + 5 = 12
Number of ways to select 3 blue balls = 5C3 = 10
Number of ways to select 3 balls from the bag = 12C3 = 220
Therefore, the probability of selecting all 3 blue balls = 10/220 = 1/22
A factory produces light bulbs, and 10% of the bulbs are defective. A sample of 20 bulbs is randomly selected. What is the probability that at most 2 bulbs are defective?
Solution:
Number of non-defective bulbs in the sample = 20 – (20 x 0.1) = 18
Probability that all 20 bulbs are non-defective = (0.9)^20 = 0.1215766546
Probability that exactly 1 bulb is defective = (20C1 x 0.1 x (0.9)^19) = 0.2706705665
Probability that exactly 2 bulbs are defective = (20C2 x (0.1)^2 x (0.9)^18) = 0.3558612159
Therefore, the probability that at most 2 bulbs are defective = 0.1215766546 + 0.2706705665 + 0.3558612159 = 0.748108437
In a deck of 52 playing cards, what is the probability of drawing a heart or a diamond?
Solution:
Number of hearts and diamonds in the deck = 13 + 13 = 26
Total number of cards in the deck = 52
Probability of drawing a heart or a diamond = 26/52 = 1/2
A box contains 10 balls, out of which 2 are black and 8 are white. If 2 balls are drawn at random without replacement, what is the probability that both balls are white?
Solution:
Total number of balls in the box = 10
Number of ways to select 2 white balls = 8C2 = 28
Number of ways to select 2 balls from the box = 10C2 = 45
Therefore, the probability of selecting two white balls = 28/45.
