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.
Nov 17Live Online Classes with Top GRE Expert Rich Zwelling
08:00 PM EST
-10:00 PM EST
Get live instruction from a top-scoring, industry-recognized GRE teachers + 6 months of access to TTP Self-Study included! Tues/Thur Sept 7, 2026---Feb. 2, 2027, 1:00pm-3:00pm EST
Dec 14Live Online Classes with Top GRE Expert Michael Butville
08:00 PM EST
-10:00 PM EST
Get live instruction from a top-scoring, industry-recognized GRE teachers + 6 months of access to TTP Self-Study included! Mon/Wed Dec. 14, 2026---Feb. 22, 2027, 8:00pm-10:00pm EST
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and
[#permalink]
22 Aug 2021, 22:56
22 Aug 2021, 22:56
3
2
10
Bookmarks
00:00
Question Stats:
48% (02:30) correct
51% (02:10) wrong based on 74 sessions
Hide Show timer Statistics
A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and 2 torches are defective.
A person takes out two items from the bag at random.
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
A person takes out two items from the bag at random.
Quantity A |
Quantity B |
Probability that both are bulbs or both are in working condition |
\(\frac{6}{7}\) |
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
Re: A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and
[#permalink]
23 Aug 2021, 13:12
23 Aug 2021, 13:12
1
1
Bookmarks
c
The probability of selecting 2 bulbs in a row is 2/3 * 9/14 = 3/7 added to the probability of selecting 2 items in working condition which happens to be the same 10/15 * 9/14 = 3/7 and together that's 6/7.
The probability of selecting 2 bulbs in a row is 2/3 * 9/14 = 3/7 added to the probability of selecting 2 items in working condition which happens to be the same 10/15 * 9/14 = 3/7 and together that's 6/7.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and
[#permalink]
03 Sep 2021, 20:18
03 Sep 2021, 20:18
2
Explanation:
Defective bulbs = 3
Working bulbs = 7
Defective torches = 2
Working torches = 3
Picking 2 bulbs out of 10 = \(^{10}C_2 = 45\)
Picking 2 torches in working condition out of 3 = \(^3C_2 = 3\)
Picking 1 bulb and 1 torch in working condition = \((^7C_1)(^3C_1) = (7)(3) = 21\)
NOTE: Picking 2 bulbs in working condition have already been included in case 1
Total cases = \(45 + 3 + 21 = 69\)
Total number of ways of picking 2 objects out of 15 = \(^{15}C_2 = 105\)
Col. A: \(\frac{69}{105}\)
Col. B: \(\frac{6}{7}\)
Cross-Multiply the denominators;
Col. A: \(483\)
Col. B: \(630\)
Hence, option B
Defective bulbs = 3
Working bulbs = 7
Defective torches = 2
Working torches = 3
Picking 2 bulbs out of 10 = \(^{10}C_2 = 45\)
Picking 2 torches in working condition out of 3 = \(^3C_2 = 3\)
Picking 1 bulb and 1 torch in working condition = \((^7C_1)(^3C_1) = (7)(3) = 21\)
NOTE: Picking 2 bulbs in working condition have already been included in case 1
Total cases = \(45 + 3 + 21 = 69\)
Total number of ways of picking 2 objects out of 15 = \(^{15}C_2 = 105\)
Col. A: \(\frac{69}{105}\)
Col. B: \(\frac{6}{7}\)
Cross-Multiply the denominators;
Col. A: \(483\)
Col. B: \(630\)
Hence, option B
