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.
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 = 117 = 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 nr = 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 31 ways and 6 non-green marbles in 86 ways. Thus, total number of ways = 31*86 = [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 32 ways. We can choose 5 non-green marbles from 8 non-green marbles in 85 ways.
Thus, the total number of ways = 32*85 = [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 33 ways. We can choose 4 non-green marbles from 8 non-green marbles in 84 ways.
Thus, the total number of ways = 33*84 = 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 = 95 = 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 22 ways. She will choose 1 red marble from 3 red marbles in 31 ways. She will choose 1 white marble from 3 white marbles in 31 ways. She will choose 1 purple marble from 1 purple marble in 11 ways.
Thus, the total number of ways = 22*31*31*11 = 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.
Suzan sees a bag of marbles, she grabs a handful at random. She has seen a...
1. Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, two green ones, five 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. HINT [See Example 1.] She has all the red ones. 2. Recall from Example 1 that whenever Suzan sees a bag of marbles, she...
Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing two red marbles, three green ones, four white ones, and two purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She does not have all the red ones. Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a...
1. Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, four green ones, three white ones, and two purple ones. She grabs seven of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT [See Example 1.] She has all the red ones. 2. Recall from Example 1 that whenever Suzan sees a bag of marbles,...
Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, four green ones, three white ones, and two purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She has at least one green one.
Recall from Example 1 that whenever 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 five of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She has two red ones and one of each of the other colors.
Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, two green ones, five white ones, and three purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She has all the red ones. I
Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, five green ones, two white ones, and two purple ones. She grabs five of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT [See Example 1.] She has at most one green one.
1. + -/1 points WaneFM7 7.4.001. My Notes Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, four green ones, three white ones, and three purple ones. She grabs seven of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT (See Example 1. She has all the red ones. Need Help? Read It Watch...
2/3 Points] DETAILS PREVIOUS ANSWERS WANEFMAC7 8.4.004.NVA MY NOTES PRACTICE ANOTHER Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, three green ones, four white ones, and one purple one. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT (See Example 1.] She has at least one green one. 114...
answer the most q's possible thank you so much 1. 0/1 points Previous Answers WaneFMAC78.4.001 My Notes Ask Your Teacher Recall from Example 1 that whenever Suran sees a bag of marbles, she grabs a handful at random. She has seen a bag containing the red marbles, five green ones, two white ones, and three purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT(See Example 1.)...