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A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and
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22 Aug 2021, 22:56
22 Aug 2021, 22:56
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Question Stats:
49% (02:42) correct
50% (02:07) wrong based on 51 sessions
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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
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23 Aug 2021, 13:12
23 Aug 2021, 13:12
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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
