5. A bag contains 7 green marbles and 4 white marbles. You select a marble at random. What are the odds in favor of picking a green marble?
A. 7:11**?
B. 7:4
C. 4:7
D. 3:7 ------------------------------------ 6. Food Express is running a special promotion in which customers can win a free gallon of milk with their food purchase if there is a star on their receipt. So far, 147 of the first 156 customers have not received a star on their receipts. What is the experimental probability of winning a free gallon of milk?
A. 11/156
B. 49/52
C. 2/39
D. 3/52
7. A bag contains 7 green marbles, 9 red marbles, 10 orange marbles, 5 brown marbles, and 10 blue marbles. You choose a marble, replace it, and choose again. Find P(red, then blue).
A. 77/164
B. 19/41*
C. 90/1681
D. 45/41
#5. Nope
The probability is 7/11, but the odds are 7:4
That is, 7 chances for and 4 against.
#6
There are 147 without a star, so 9 with a star. So
P(star) = 9/156 = 3/52
#7
total marbles: 41
P(red) = 9/41
P(blue) = 10/41
P(red,blue) = 9/41 * 10/41 = 90/1681
You can't just add the probabilities. If so, then by specifying enough drawings, you'd eventually add up enough fractions to get more than 1. Actually, by specifying more and more drawings, the chance of getting them ALL keeps getting smaller.
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The odds in favor of picking a green marble would be given by:
Number of green marbles : Number of non-green marbles = 7:4
So the odds in favor of picking a green marble are 7 to 4.
The probability of choosing a red marble on the first draw is 8/43, because there are 8 red marbles in a total of 43 marbles. After replacing this marble back into the bag, there are still 12 blue marbles left in a total of 43 marbles. Therefore, the probability of choosing a blue marble on the second draw, given that a red marble was chosen on the first draw, is also 12/43.
Using the multiplication rule of probability, we can find the probability of both events occurring as:
P(red, then blue) = P(red) x P(blue | red)
P(red, then blue) = (8/43) x (12/43)
P(red, then blue) = 96/1849
So the probability of choosing a red marble on the first draw and a blue marble on the second draw is 96/1849.
agreed
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That doesn't make sense. Please let me know if you have any questions or if there's anything I can help you with.
A bag contains 7 green marbles and 4 white marbles. You select a marble at random. What are the odds in favor of picking a green marble?
A bag contains 5 green marbles, 8 red marbles, 11 orange marbles, 7 brown marbles, and 12 blue marbles. You choose a marble, replace it, and choose again. What is P(red, then blue)?
You own 5 pairs of jeans and want to take 2 of them with you on vacation. In how many ways can you choose 2 pairs of jeans?
You can choose 2 pairs of jeans out of 5 in C(5,2) ways, where C(n,r) represents the number of combinations of n objects taken r at a time.
C(5,2) = (5!)/(2!(5-2)!) = (5x4)/(2x1) = 10
So, you can choose 2 pairs of jeans out of 5 in 10 ways.
You and 5 friends go to a concert. In how many different ways can you sit in the assigned seats?
There are 6 people who need to be seated. The first person can sit in any of the 6 seats. Once the first person has chosen a seat, there are only 5 seats left for the second person to choose from. Similarly, there will be 4 seats left for the third person, 3 seats left for the fourth person, 2 seats left for the fifth person, and only 1 seat left for the last person.
Therefore, the total number of different ways that the group of 6 people can be seated in the assigned seats is:
6 x 5 x 4 x 3 x 2 x 1 = 720
So, there are 720 different ways for the group of 6 people to be seated in their assigned seats.
8. Each of two urns contains green balls and red balls. Urn 1 contains 10 green balls and 14 red balls. Urn 2 contains 4 green balls and 11 red balls. If a ball is drawn from each urn, what is P(red to red)?
The probability of drawing a red ball from urn 1 is 14/(10+14) = 7/12.
The probability of drawing a red ball from urn 2 is 11/(4+11) = 11/15.
The probability of drawing a red ball from urn 1 and a red ball from urn 2 is the product of these two probabilities:
P(red to red) = P(red from urn 1) * P(red from urn 2)
P(red to red) = (7/12) * (11/15)
P(red to red) = 77/180
So the probability of drawing a red ball from each urn is 77/180.
You have four $1 bills, two $5 bills, five $10 bills, and five $20 bills in your wallet. You select
a bill at random. Without replacing the bill, you choose a second bill. What is P($1, then $10)?
The probability of selecting a $1 bill on the first draw is 4/16, because there are 4 $1 bills in a total of 16 bills. After removing the selected $1 bill, there are now 15 bills remaining, with 5 of them being $10 bills. Therefore, the probability of selecting a $10 bill on the second draw, given that a $1 bill was chosen on the first draw, is 5/15.
Using the multiplication rule of probability, we can find the probability of both events occurring as:
P($1, then $10) = P($1) x P($10 | $1)
P($1, then $10) = (4/16) x (5/15)
P($1, then $10) = 5/48
So the probability of selecting a $1 bill on the first draw and a $10 bill on the second draw is 5/48.
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5
peaches. Jameson picks a fruit at random and does not replace it. Then Brittany picks a fruit at
random. What is the probability that Jameson gets a banana and Brittany gets a pear?
The probability of Jameson selecting a banana on the first draw is 2/14, because there are 2 bananas in a total of 14 pieces of fruit. After Jameson selects a fruit, there are now only 13 pieces of fruit remaining, with only 1 pear left. Therefore, the probability of Brittany selecting a pear on the second draw, given that Jameson selected a banana on the first draw, is 1/13.
Using the multiplication rule of probability, we can find the probability of both events occurring as:
P(banana, then pear) = P(banana) x P(pear | banana)
P(banana, then pear) = (2/14) x (1/13)
P(banana, then pear) = 1/91
So the probability of Jameson selecting a banana on the first draw and Brittany selecting a pear on the second draw is 1/91.
The probability of a certain baseball player hitting a foul ball is . How many foul balls
would you expect her to hit after 80 swings?
If the probability of hitting a foul ball is p = 0.25, then the expected number of foul balls hit after n swings is given by the formula:
Expected number = p * n
In this case, we are given that n = 80 swings. Therefore, the expected number of foul balls hit is:
Expected number = 0.25 * 80
Expected number = 20
So, we would expect the player to hit 20 foul balls after 80 swings.
A true-false test has 5 questions. What is the probability of guessing the correct answers to all
of the questions?
The probability of guessing any one of the five true-false questions correctly by random guessing is 1/2, since there are two possible answers (true or false) and only one is correct.
Therefore, the probability of guessing all five questions correctly is:
P(all correct) = (1/2)^5 = 1/32
So the probability of guessing all of the answers correctly is 1/32.
You and 3 friends go to a concert. In how many different ways can you sit in the assigned
seats?
There are 4 people who need to be seated. The first person can sit in any of the 4 seats. Once the first person has chosen a seat, there are only 3 seats left for the second person to choose from. Similarly, there will be 2 seats left for the third person, and only 1 seat left for the last person.
Therefore, the total number of different ways that the group of 4 people can be seated in the assigned seats is:
4 x 3 x 2 x 1 = 24
So, there are 24 different ways for the group of 4 people to be seated in their assigned seats.
You own 5 pairs of jeans and want to take 2 of them with you on vacation. In how many ways
can you choose 2 pairs of jeans?
You can choose 2 pairs of jeans out of 5 in C(5,2) ways, where C(n,r) represents the number of combinations of n objects taken r at a time.
C(5,2) = (5!)/(2!(5-2)!) = (5x4)/(2x1) = 10
So, you can choose 2 pairs of jeans out of 5 in 10 ways.
A box contains 95 pink rubber bands and 90 brown rubber bands. You select a rubber band at random from the box. Find each probability. Write the probability as a fraction in simplest form.
Find the theoretical probability of selecting a pink rubber band.
Find the theoretical probability of selecting a brown rubber band.
You repeatedly choose a rubber band from the box, record the color, and put the rubber band back in the box. The results are shown in the table below. Find the experimental probability of each color based on the table.
Outcome Occurrences
Pink 36
Brown 33
Theoretical probability of selecting a pink rubber band = (Number of pink rubber bands in the box) / (Total number of rubber bands in the box) = 95 / (95 + 90) = 19/37
Theoretical probability of selecting a brown rubber band = (Number of brown rubber bands in the box) / (Total number of rubber bands in the box) = 90 / (95 + 90) = 18/37
Experimental probability of selecting a pink rubber band = Number of occurrences of selecting a pink rubber band / Total number of selections = 36 / (36 + 33) = 12/23
Experimental probability of selecting a brown rubber band = Number of occurrences of selecting a brown rubber band / Total number of selections = 33 / (36 + 33) = 11/23