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

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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

The official solution has been posted. Looking forward to a healthy discussion..

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 torch-bulb and in other bulb-torch..
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..

Thanks chetan2u this helps.

Hi chetan2u

Can 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??

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!!!

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!(10-2)!] = (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!(15-2)!] = (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!(10-2)!] = (10 x 9)/2! = 45. The total number of ways to select 2 items from the group of 15 items is 15C2 = 15!/[2!(15-2)!] = (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!(7-2)!] = (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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