A fun size bag of M&Ms has 4 blue, 3 orange, 3 red, 2 green, 2...

A fun size bag of M&Ms has 4 blue, 3 orange, 3 red, 2 green, 2 yellow, and 1 brown M&Ms. What is the probability of randomly selecting 5 M&Ms where 3 are blue and 2 are orange?

Solutions

Expert Solution

Number of blue M&Ms in a fun size bag = 4

Number of orange M&Ms in a fun size bag = 3

Number of red M&Ms in a fun size bag = 3

Number of green M&Ms in a fun size bag = 2

Number of yellow M&Ms in a fun size bag = 2

Number of brown M&Ms in a fun size bag = 1

Total number of M&Ms in a fun size bag = 4+3+3+2+2+1=15

Probability of randomly selecting 5 M&Ms where 3 are blue and 2 are orange

= (Number of ways of selecting 3 blue M&Ms from 4 blue M&Ms x Number of ways of selecting 2 orange M&Ms from 3) / Number of ways of selecting 5 M&Ms from a total of 15 M&Ms

Number of ways of selecting 3 blue M&Ms from 4 blue M&Ms = =4

Number of ways of selecting 2 orange M&Ms from 3 = =3

Number of ways of selecting 5 M&Ms from a total of 15 M&Ms==3003

Probability of randomly selecting 5 M&Ms where 3 are blue and 2 are orange

= (Number of ways of selecting 3 blue M&Ms from 4 blue M&Ms x Number of ways of selecting 2 orange M&Ms from 3) / Number of ways of selecting 5 M&Ms from a total of 15 M&Ms

= (4 x 3)/3003 = 12/3003 = 0.003996004

Probability of randomly selecting 5 M&Ms where 3 are blue and 2 are orange = 0.003996004

milcah answered 2 hours ago

Next > < Previous

Related Solutions

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue,...

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement. Determine the probability of the following. (Enter your probabilities as fractions.) -The first candy drawn is red. -The second candy drawn is red, given that the first candy drawn is red. -The third candy drawn is green, given that the first two candies drawn...

A bag contains 4 Blue, 5 red and 3 green balls. Debolina is asked to pick...

A bag contains 4 Blue, 5 red and 3 green balls. Debolina is asked to pick up three balls at random, one by one without replacement from that bag. If the balls picked up are of same colour then Debolina will get Rs. 7000, if the balls picked up are of different colour, then she will get Rs. 9000, but otherwise she has to pay Rs. 4000. a) Find the expected gain of Debolina if she plays the game. b)...

1. Let M denote set of m&ms in a bag. This consist of red, yellow, green,...

1. Let M denote set of m&ms in a bag. This consist of red, yellow, green, blue and brown candies. a. Device on equivalence relation on M. b. Define relation R on M by: aRb if and only if either: a is green or b is blue, or, a is yellow abd b is not yellow. Is R symmetric, reflexive, transitive or anti-symmetric? Explain. 2. On set Zx(Z not including {0}) define relation R (a,b) a,b e Z, b#0 as...

Suppose you have a bag of M&Ms with 19 M&Ms. Suppose 4 of them are red,...

Suppose you have a bag of M&Ms with 19 M&Ms. Suppose 4 of them are red, 3 are green, and 12 are yellow. (a) If one M&M is chosen at random from the bag, find the probability that it is yellow. (b) If one M&M is chosen at random from the bag, and eaten, and then a second M&M is chosen at random from the bag, find the probability that they are both red.

A jar contains 4 Orange, 3 Blue, and 2 Green marbles. (a) Two marbles are selected...

A jar contains 4 Orange, 3 Blue, and 2 Green marbles. (a) Two marbles are selected one at a time at random without replacement, so order is observed in the sample. Define the events:A=“the first marble is Orange”, B= “the second marble is Orange”. Find(i)P(A), (ii)P(B), (iii)P(A∪B) =P(at least one Orange marble is obtained in the 2 draws). Show your work. (b) Suppose instead three marbles are chosen at random from the jar by choosing 3 in one hand; no...

The distribution of M&M colors is 24% blue, 20% orange, 16% green, 14% yellow, 13% red,...

The distribution of M&M colors is 24% blue, 20% orange, 16% green, 14% yellow, 13% red, and 13% brown. You have a small bag of M&M's and count the colors. Your bag has 9 blue, 9 orange, 4 green, 12 yellow, 3 red, and 6 brown M&M's. Before you begin, how many M&M's do you need in your sample (minimum) to be able to do this test? Remember, your assumption for chi-square is that every expected frequency has at least...

Bag Blue Orange Green Yellow Red Brown Total Number of Candies 1 9 13 14 10...

Bag Blue Orange Green Yellow Red Brown Total Number of Candies 1 9 13 14 10 7 7 60 2 13 10 6 9 9 8 55 3 13 12 4 10 9 6 54 4 16 13 8 6 6 8 57 5 10 10 12 5 15 4 56 6 9 18 3 6 12 12 60 7 11 13 6 15 8 6 59 8 12 18 5 9 6 5 55 9 12 10 8 15...

Bag Blue Orange Green Yellow Red Brown Total Number of Candies 1 9 13 14 10...

Assume that a bag initially contains 6 balls: 2 red, 2 green and 2 blue balls....

Assume that a bag initially contains 6 balls: 2 red, 2 green and 2 blue balls. At each step, you choose a ball from the bag at random, note its color, but do not put it back into the bag. Instead, you add to the bag two balls, which are of of two different colors, and different in color from the color of the removed ball. (For example, if you choose a red ball in the first step, then after...

According to M&Ms are randomly mixed to have 24% blue M&MS. Suppose your bag has a...

According to M&Ms are randomly mixed to have 24% blue M&MS. Suppose your bag has a total of 57 M&Ms and 11 are blue. Assuming the distribution of total m&m is normally distribution, what is the mean and standard deviation of the sample distribution? Whats the standard score? Whats the null hypothesis and alternate hypothesis? ONE sided or two sided? What is the p-value At the 5% significance level, what would our conclusions be?

Subjects