Question

Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, three green ones, four white ones, and one purple one. She grabs seven of them. Find the probability of the following event, expressing it as a fraction in lowest terms.

She has at least one green one.

Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, two green ones, three white ones, and one purple one. She grabs five of them. Find the probability of the following event, expressing it as a fraction in lowest terms.

She has two green ones and one of each of the other colors.

Answer 1

First question :

Number of red marbles = 3

Number of green marbles = 3

Number of white marbles = 4

Number of purple marbles = 1

Thus, total number of marbles = 3+3+4+1 = 11

Suzan grabbed 7 marbles. Thus, total number of possible outcomes = 11$\mathbb{C}$7 = 11!/{7!(11-7)!} = 11!/(7!*4!) = (11*10*9*8*7!)/(7!*4*3*2*1) = (11*10*9*8)/(4*3*2*1) = 7920/24 = 330.[as n$\mathbb{C}$r = n!/{r!(n-r)!}, where, x! = x(x-1)(x-2).....1]

Thus, total number of possible outcomes = 330. ------(2)

We know, Probability of any event = (Number of favorable outcomes)/(Total number of possible outcomes). ---------(1)

Now, we have to find the probability that she has at least one green one. There are 3 green marbles, thus, there will be 3 cases. Thus, number of non-green marbles = 11-3 = 8.

Case 1 : 1 green marble and 6 non-green marbles - There are 3 green marbles and 8 non-green marbles. Thus, we have to select 1 green marble from 3 green marbles and 6 non-green marbles from 8 non-green marbles.

We can choose 1 green marble in 3$\mathbb{C}$1 ways and 6 non-green marbles in 8$\mathbb{C}$6 ways. Thus, total number of ways = 3$\mathbb{C}$1*8$\mathbb{C}$6 = [3!/(2!*1!)]*[8!/(6!*2!)] = [(3*2!)/(2!*1)]*[(8*7*6!)/(6!*2*1)] = 3*28 = 84 [formula is given above]

Thus, number of ways = 84.

Case 2 : 2 green marbles and 5 non-green marbles - There are 3 green marbles and 8 non-green marbles. Thus, we can choose 2 green marbles from 3 green marbles in 3$\mathbb{C}$2 ways. We can choose 5 non-green marbles from 8 non-green marbles in 8$\mathbb{C}$5 ways.

Thus, the total number of ways = 3$\mathbb{C}$2*8$\mathbb{C}$5 = [3!/(2!*1!)]*[8!/(5!*3!)] = [(3*2!)/(2!*1)]*[(8*7*6*5!)/(5!*3*2*1)] = 3*[(8*7*6)/(3*2*1)] = 3*56 = 168. Thus, total number of ways = 168.

Case 3 : 3 green marbles and 4 non-green marbles - There are 3 green marbles and 8 non-green marbles. We can choose 3 green marbles from 3 green marbles in 3$\mathbb{C}$3 ways. We can choose 4 non-green marbles from 8 non-green marbles in 8$\mathbb{C}$4 ways.

Thus, the total number of ways = 3$\mathbb{C}$3*8$\mathbb{C}$4 = 1*[8!/(4!*4!)] = (8*7*6*5*4!)/(4!*4*3*2*1) = (8*7*6*5)/(4*3*2*1) = 1680/24 = 70. Thus, total number of ways = 70.

Thus, adding all the three cases, the number of favorable outcomes = 84+168+70 = 322.

Thus, from (1) and (2), the required probability = 322/330 = 161/165(reduced to lowest fraction).

Thus, the probability that she has at least one green marble = 161/165 .

Second question :

Number of red marbles = 3

Number of green marbles = 2

Number of white marbles = 3

Number of purple marbles = 1

Thus, total number of marbles = 3+2+3+1 = 9.

Suzan grabs 5 of them. Thus, Suzan grabs 5 marbles from the 9 marbles. Thus, total number of possible outcomes = 9$\mathbb{C}$5 = 9!/(5!*4!) = (9*8*7*6*5!)/(5!*4*3*2*1) = (9*8*7*6)/(4*3*2*1) = 3024/24 = 126. Thus, total number of possible outcomes = 126. -----------(3)

We have to find the probability that she has two green ones and one of each colour. Thus, she will choose 2 green marbles from 2 green marbles in 2$\mathbb{C}$2 ways. She will choose 1 red marble from 3 red marbles in 3$\mathbb{C}$1 ways. She will choose 1 white marble from 3 white marbles in 3$\mathbb{C}$1 ways. She will choose 1 purple marble from 1 purple marble in 1$\mathbb{C}$1 ways.

Thus, the total number of ways = 2$\mathbb{C}$2*3$\mathbb{C}$1*3$\mathbb{C}$1*1$\mathbb{C}$1 = 1*3*3*1 = 9 [using the formula written above].

Thus, number of favorable outcomes = 9.

Thus, from (1) and (3), we have the required probability = 9/126 = 1/14(reduced to lowest fraction).

Thus, the probability that she has two green ones and one of each of the other colours = 1/14.