GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
A bag contains 12 marbles: 5 of the marbles are red, 3 are g
[#permalink]
22 Nov 2017, 19:07
22 Nov 2017, 19:07
Expert Reply
7
Bookmarks
00:00
Question Stats:
36% (01:42) correct
63% (01:33) wrong based on 148 sessions
Hide Show timer Statistics
A bag contains 12 marbles: 5 of the marbles are red, 3 are green, and the rest are blue.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The probability of consecutively choosing two red marbles and a green marble without replacement |
The probability of consecutively choosing a red and two blue marbles with replacement |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Drill 2
Question: 5
Page: 525
Question: 5
Page: 525
Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g
[#permalink]
24 Nov 2017, 06:21
24 Nov 2017, 06:21
2
qty A = (5/12 * 4/11 * 3/10) => Without replacement
qty B = 5/12* 4/12* 3/12) => with Replacement
Therefore B>A
qty B = 5/12* 4/12* 3/12) => with Replacement
Therefore B>A
Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g
[#permalink]
29 Dec 2017, 13:25
29 Dec 2017, 13:25
2
Expert Reply
1
Bookmarks
Explanation
Quantity A asks for the probability “without replacement,” so that means you have to take into account that there will be one marble less in the total after each draw.
The probability of first choosing a red marble is \(\frac{5}{12}\), a second red marble is \(\frac{4}{11}\), and then a green marble is \(\frac{3}{10}\).
This is an “and” probability problem, so you have to multiply the probability of each event together: \(\frac{5}{12}\times\frac{4}{11}\times\frac{3}{10}=\frac{60}{1320}=\frac{1}{22}\).
For Quantity B, you do the same thing, but the total stays the same for each draw: \(\frac{5}{12}\times\frac{4}{12}\times\frac{4}{12}=\frac{80}{1728}=\frac{5}{108}\).
Quantity B is greater.
Quantity A asks for the probability “without replacement,” so that means you have to take into account that there will be one marble less in the total after each draw.
The probability of first choosing a red marble is \(\frac{5}{12}\), a second red marble is \(\frac{4}{11}\), and then a green marble is \(\frac{3}{10}\).
This is an “and” probability problem, so you have to multiply the probability of each event together: \(\frac{5}{12}\times\frac{4}{11}\times\frac{3}{10}=\frac{60}{1320}=\frac{1}{22}\).
For Quantity B, you do the same thing, but the total stays the same for each draw: \(\frac{5}{12}\times\frac{4}{12}\times\frac{4}{12}=\frac{80}{1728}=\frac{5}{108}\).
Quantity B is greater.
