Latest Wipro Aptitude Question SOLUTION: A, B ans C are three speakers. They have to speak randomly along with another 5 speakers in a function. A has to speak before B and B has to speak before C. What is the probability

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A, B ans C are three speakers. They have to speak randomly along with another 5 speakers in a function.
A has to speak before B and B has to speak before C. What is the probability???


a) 1/4
b) 1/6
c) 1/8
d) 1/10

Asked In Wipro SATHISH T.N (11 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

Aman  (11 years ago)

1/6

Without considering other speakers , their arrangement should be ABC
whereas 6 arrangements are possible among these 3 persons.

so 1/6 probability

Is this solution Helpfull? Yes (9) | No (9)  

sobhna  (11 years ago)

But question has given they have to speak randomly with another 5 speakers what about it /????

Is this solution Helpfull? Yes (3) | No (3)  

sumaja chaganti  (11 years ago)

let us consider abc as 1 group.. then among with other 5 speakers the total ways are 5+1group =6! ways *3!(abc will arrange themselves). so probability is 6!/6!*3!=1/3!=1/6

Is this solution Helpfull? Yes (1) | No   

sarthak  (8 years ago)

To take a simple example: suppose there are 4 people, A, B, C, D speaking. What is the probability that A, B, C speak in the order ABC? There are, of course, 4!= 24 different ways to order 4 things: ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD BCDA BDAC BDCA CBAD CBDA CABD CADB CDBA CDAB DBCA DBAC DCBA DCAB DABC DACB Of those, ABCD, ABDA, ADBC, DABC have A, B, C in that order: the probability that A, B, C will speak in that order is 4/24= 1/6. If instead we count only A, B, C, we find 3!= 6 different orders and only 1 (ABC) has them in that order: Again 1/6. Given any n people speaking in random order, the probability that a given 3 will speak in a given order is 1!/3!= 1/6. If you use all orders for n people as the denominator and all orders for n people that have the same 3 people in the given order, the numerator and denominator are both multiplied by n!/3!.

Is this solution Helpfull? Yes (1) | No   

mounika  (11 years ago)

can any one plz explain clearly?

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harshavi  (9 years ago)

but i think according to the que we could not take abc as one group since they have given that abc speaks randomly with others

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birbal ram sain  (6 years ago)

assume abc is a group.so total number of ways arragment is (abc as a 1 plus 5 other) 6! .6! is favorable outcomes and 8! is total outcomes.so probability is 6!/8!.

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PIUSH SINGH  (6 years ago)

The total number of ways in which 8 person can speak is -: 8P8=8!
The number of ways in which A,B and C can be arranged in the specified speaking order is :-8C3
There are always 5! ways in which the other five can speak.
so the favourable no of ways is
8C3*5!/8!=1/6

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v.pravalika  (6 years ago)

A probability of speaking out of 8 is 1/8
B probability of speaking out of 7 is 1/7
c probability of speaking out of 6 is 1/6
1/8+1/7+1/6=1/8

Is this solution Helpfull? Yes | No (1)  

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