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A bag contains 10 bulbs and 5 torches, out of which 3 bu
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27 Apr 2017, 22:48
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A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and 2 torches are defective. If a person takes out two items from the bag at random, what is the probability that both are bulbs or both are in working condition? A. 8/35 B. 3/7 C. 23/35 D. 6/7 E. None of these Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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A bag contains 10 bulbs and 5 torches, out of which 3 bu
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Updated on: 01 May 2017, 06:43
The official solution has been posted. Looking forward to a healthy discussion..
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A bag contains 10 bulbs and 5 torches, out of which 3 bu
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Updated on: 28 Apr 2017, 10:55
Probability that both are bulbs:10/15 * 9/14 = 3/7 Probability that both are working: 1 working bulb & 1 working Torch = 7/15 * 3/14 = 1/10 2 Working torches = 3/15 * 2/14 = 1/5 * 1/7 = 1/35 Total probability = 3/7 + 1/10 + 1/35 = 30+7+2/70 = 39/70 chetan2u, thanks for pointing out. I read the question wrong. When I tried again, I still do not get the correct answer. Could you please let me know where my approach is going wrong?
Originally posted by quantumliner on 28 Apr 2017, 09:03.
Last edited by quantumliner on 28 Apr 2017, 10:55, edited 1 time in total.



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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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28 Apr 2017, 10:05
quantumliner wrote: The probability of selecting the first item as a bulb in working condition = Total number of working bulbs/Total number of items = 7/15
The probability of selecting the second item as a bulb in working condition = Total number of remaining working bulbs/Total number of remaining items = 6/14
Total probability = 7/15 * 6/14 = 1/5
Answer is E. None of these Relook into your solution... EgmatQuantExpert wrote: A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and 2 torches are defective. If a person takes out two items from the bag at random, what is the probability that both are bulbs or both are in working condition?
A. 8/35 B. 3/7 C. 23/35 D. 6/7 E. None of these There are three categories we have to look for ... 1) choosing 2 bulbs out of 10..... This will cater for 2 bulbs in working condition.. 10C2=\(\frac{10*9}{2}\)= 45.. 2) choosing 2 torches in working condition.. 3C2=3.. 3) choosing one bulb out of 7 working and one torch out of 3 working.. 7*3=21.. Total ways under the given scenario=45+3+21=69.. Total ways possible=(\(10+5)C2=15C2=\frac{15*14}{2}=105\).. Probability=\(\frac{69}{105}=\frac{23}{35}\) C
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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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28 Apr 2017, 12:17
quantumliner wrote: Probability that both are bulbs:10/15 * 9/14 = 3/7 Probability that both are working: 1 working bulb & 1 working Torch = 7/15 * 3/14 = 1/10 2 Working torches = 3/15 * 2/14 = 1/5 * 1/7 = 1/35 Total probability = 3/7 + 1/10 + 1/35 = 30+7+2/70 = 39/70 chetan2u, thanks for pointing out. I read the question wrong. When I tried again, I still do not get the correct answer. Could you please let me know where my approach is going wrong? You are wrong in coloured portion.. In the way you are taking your denominator and other cases, you are talking of permutation that is the arrangement too matters.. So in one case you pick torchbulb and in other bulbtorch.. So ways become 7/15*3/14+3/15*7/14=2/10=1/5.. Ans = 3/7+1/5+1/35=(30+14+2)/70=46/70=23/35.. Hope it helps..
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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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28 Apr 2017, 14:03
Thanks chetan2u this helps.



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A bag contains 10 bulbs and 5 torches, out of which 3 bu
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01 May 2017, 03:32
chetan2u wrote: There are three categories we have to look for ... 1) choosing 2 bulbs out of 10..... This will cater for 2 bulbs in working condition.. 10C2=\(\frac{10*9}{2}\)= 45.. 2) choosing 2 torches in working condition.. 3C2=3.. 3) choosing one bulb out of 7 working and one torch out of 3 working.. 7*3=21..
Total ways under the given scenario=45+3+21=69.. Total ways possible=(\(10+5)C2=15C2=\frac{15*14}{2}=105\)..
Probability=\(\frac{69}{105}=\frac{23}{35}\)
C
Hi chetan2uCan you please explain why are we considering only favorable number of items i.e. 10 or 3 or 7 and not 15 for all cases??



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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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01 May 2017, 03:58
chetan2u wrote: quantumliner wrote: The probability of selecting the first item as a bulb in working condition = Total number of working bulbs/Total number of items = 7/15
The probability of selecting the second item as a bulb in working condition = Total number of remaining working bulbs/Total number of remaining items = 6/14
Total probability = 7/15 * 6/14 = 1/5
Answer is E. None of these Relook into your solution... EgmatQuantExpert wrote: A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and 2 torches are defective. If a person takes out two items from the bag at random, what is the probability that both are bulbs or both are in working condition?
A. 8/35 B. 3/7 C. 23/35 D. 6/7 E. None of these There are three categories we have to look for ... 1) choosing 2 bulbs out of 10..... This will cater for 2 bulbs in working condition.. 10C2=\(\frac{10*9}{2}\)= 45.. 2) choosing 2 torches in working condition.. 3C2=3.. 3) choosing one bulb out of 7 working and one torch out of 3 working.. 7*3=21.. Total ways under the given scenario=45+3+21=69.. Total ways possible=(\(10+5)C2=15C2=\frac{15*14}{2}=105\).. Probability=\(\frac{69}{105}=\frac{23}{35}\) C Hi, The 2nd condition mentions both in working condition and does not specify a working bulb or a torch....so why can't we consider 10 working items and work that way...so 10C2?



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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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01 May 2017, 04:08
salonipatil wrote: chetan2u wrote: quantumliner wrote: The probability of selecting the first item as a bulb in working condition = Total number of working bulbs/Total number of items = 7/15
The probability of selecting the second item as a bulb in working condition = Total number of remaining working bulbs/Total number of remaining items = 6/14
Total probability = 7/15 * 6/14 = 1/5
Answer is E. None of these Relook into your solution... A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and 2 torches are defective. If a person takes out two items from the bag at random, what is the probability that both are bulbs or both are in working condition? A. 8/35 B. 3/7 C. 23/35 D. 6/7 E. None of these There are three categories we have to look for ... 1) choosing 2 bulbs out of 10..... This will cater for 2 bulbs in working condition.. 10C2=\(\frac{10*9}{2}\)= 45.. 2) choosing 2 torches in working condition.. 3C2=3.. 3) choosing one bulb out of 7 working and one torch out of 3 working.. 7*3=21.. Total ways under the given scenario=45+3+21=69.. Total ways possible=(\(10+5)C2=15C2=\frac{15*14}{2}=105\).. Probability=\(\frac{69}{105}=\frac{23}{35}\) C Hi, The 2nd condition mentions both in working condition and does not specify a working bulb or a torch....so why can't we consider 10 working items and work that way...so 10C2? Two mistakes in your suggestion 1) Case 1 has already considered two working bulbs so with your method that case will be counted twice 2) When you take case one one torch and one bulb then cases are counted as 7C1*3C1 instead of 10C2 cause the change of group of good bulb and torch will be counted differently I hope this helps!!!
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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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01 May 2017, 05:24
I am quite happy with the discussion so far for this question. While posting this question, I was expecting a few of you might fall into the trap and make mistake in either – 1. Considering Permutation instead of just Combination 2. Making a mistake while choosing the two items – a few of you might not consider all the working condition cases or miss out on subtracting a few additional cases.
Let me explain the way I solve this question – • We need to choose 2 items from 15 items, that can be done in \(^{15}C_2\) ways \(= 15*\frac{14}{2} = 15*7= 105\) ways.
• We need to find the probability that
o Either both are bulbs –(Case 1)
That can be chosen in \(^{10}C_2\) ways \(= 10*\frac{9}{2} = 45\) ways
Now please understand that in these 45 ways, we have considered all type of cases – both the bulbs working OR 1 bulb working and 1 being defective OR both are defective since in this case, we were choosing the bulbs without any restrictions.
So again just for everyone’s clarity
45 ways consists of – [both bulbs working OR 1 working 1defective OR both defective] o Now that we have considered the cases of both are bulbs, lets us focus on both items are in the working condition.(Case 2)
Now there are total (103) =7 bulbs and (52) = 3 torches which are working
That means total (7+3) = 10 items are working.
We need to choose 2 of them, we can do that in \(^{10}C_2\) ways \(= 45\) ways. [ 1 bulb+1 torch OR both bulbs OR both torch]
o Now focus on the bold part in Case 1, we are choosing both working bulbs in case 1 also and case 2 also, we are double counting it, so we need to subtract it once Out of 7 bulbs, we can choose 2 working bulbs in \(^7C_2\)ways = 21 ways. • Therefore, now we can say, the total ways of choosing either both bulbs or both in working condition are = 45 + 45 – 21 = 69 ways.
• Hence the probability \(= \frac{69}{105} = \frac{23}{35}\) Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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02 May 2017, 16:55
EgmatQuantExpert wrote: A bag contains 10 bulbs and 5 torches, out of which 3 bulbs and 2 torches are defective. If a person takes out two items from the bag at random, what is the probability that both are bulbs or both are in working condition?
A. 8/35 B. 3/7 C. 23/35 D. 6/7 E. None of these If we let B = both items are bulbs and W = both items are in working condition, then: Pr(B or W) = Pr(B) + Pr(W)  Pr(B and W) Let’s first determine Pr(B), the probability that both items are bulbs. Since there are 10 bulbs and we must select 2, the number of ways to select the 2 bulbs is 10C2 = 10!/[2!(102)!] = (10 x 9)/2! = 5 x 9 = 45. The total number of ways to select 2 items from the group of 15 items is 15C2 = 15!/[2!(152)!] = (15 x 14)/2! = 15 x 7 = 105. Thus, the probability of selecting 2 bulbs is 45/105 = 9/21 = 3/7. Next let’s determine Pr(W), the probability that both items are in working condition. Since there are a total of 15 items and 5 are defective, we have 15  5 = 10 items in working condition. Thus, the number of ways to select 2 working items is 10C2 = 10!/[2!(102)!] = (10 x 9)/2! = 45. The total number of ways to select 2 items from the group of 15 items is 15C2 = 15!/[2!(152)!] = (15 x 14)/2! = 15 x 7 = 105. Thus, the probability of selecting 2 working items is 45/105 = 9/21 = 3/7. Finally, let’s determine Pr(B and W), the probability of selecting 2 working bulbs. Since there are 7 working bulbs and we must select 2 of them, the number of ways to select those 2 bulbs is 7C2 = 7!/[2!(72)!] = (7 x 6)/2! = 21. Since the total number of outcomes is 105, the probability of selecting 2 working bulbs is 21/105 = 1/5. Thus, the probability of selecting 2 bulbs or 2 working items is: 3/7 + 3/7  1/5 = 6/7  1/5 = 30/35  7/35 = 23/35 Answer: C
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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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13 Jul 2017, 19:31
This is how I process this problem: We have: WorkingDefectiveTotal Bulb7310 Torch325 Total10515
As you can see, there is an overlap between bulbs and working, which should be excluded. Thus, P = P (both are bulbs) + P (both working)  P (both bulbs AND working).
1. P (both are bulbs): 10/15*9/14=3/7
2. P (both working): 10/15*9/14=3/7
3. P (both bulbs AND working): 7/15*6/14=1/5
P = 3/7+3/71/5=23/35
It is similar to what has been presented by others, however, I just thought setting up a table makes is easier to analyze this problem.



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Re: A bag contains 10 bulbs and 5 torches, out of which 3 bu
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