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25 May 2023, 02:50
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D
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There are 7 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent?
(A) 1/8
(B) 1/6
(C) 1/5
(D) 1/4
(E) 1/3
(A) 1/8
(B) 1/6
(C) 1/5
(D) 1/4
(E) 1/3
26 Jun 2023, 09:40
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Let the keys be name as A B C D E F G H I [H and I being the 2 new keys]
Probability of H and I being together = \(\frac{Total number of arrangements where H and I are together}{Total number of possible arrangements}\)
Total number of arrangements possible in a circle with the 9 keys = (n-1)! = 8!
Total number of arrangements where H and I are together = 7! x 2 [Consider H and I as one unit. In addition, there are 7 other units i.e. A B C D E F G. Total number of ways to arrange the 8 units in a circle is (n-1)! = 7!. Further, H and I can be arranged in 2 ways i.e. H I or I H.]
Probability = (7! x 2)/8! = 1/4 Ans D.
Probability of H and I being together = \(\frac{Total number of arrangements where H and I are together}{Total number of possible arrangements}\)
Total number of arrangements possible in a circle with the 9 keys = (n-1)! = 8!
Total number of arrangements where H and I are together = 7! x 2 [Consider H and I as one unit. In addition, there are 7 other units i.e. A B C D E F G. Total number of ways to arrange the 8 units in a circle is (n-1)! = 7!. Further, H and I can be arranged in 2 ways i.e. H I or I H.]
Probability = (7! x 2)/8! = 1/4 Ans D.
01 Jul 2023, 12:20
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Okay so a decent question, which has been approached by fellow member amanbachhawat. However, i'll try to approach it in a different way. So here goes:
image.jpeg [ 181.45 KiB | Viewed 2025 times ]
Have a look at attached image (marked as spoiler). In this 7 keys have been marked in a circular manner as asked in question. And for your convenience, i have already inserted a key, shown by text next to it. The position of key does not affect our case and that's the beauty of circular arrangement.
Now observe, that after inserting a key, there are 8 options available for new key to be inserted. Out of these, position 1 & 2 will be adjacent to Newly inserted key and rest i.e. position 3 to 8 will be not adjacent. Therefore,
P(adjacent) = Favorable cases/ Total cases
= 2/8
= 1/4
Option D
Attachment:
image.jpeg [ 181.45 KiB | Viewed 2025 times ]
Now observe, that after inserting a key, there are 8 options available for new key to be inserted. Out of these, position 1 & 2 will be adjacent to Newly inserted key and rest i.e. position 3 to 8 will be not adjacent. Therefore,
P(adjacent) = Favorable cases/ Total cases
= 2/8
= 1/4
Option D
Bunuel
