Probability
Aptitude

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Q.

Four boys and three girls stand in queue for an interview. The probability that they stand in alternate positions is:

 A.

1/35

 B.

1/34

 C.

1/17

 D.

1/68

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Solution:
Option(A) is correct

Total number of possible arrangements for 4 boys and 3 girls in a queue $= 7!$

When they occupy alternate position the arrangement would be like:

$B$ $G$ $B$ $G$ $B$ $G$ $B$

Thus, total number of possible arrangements for boys $= (4×3×2)$

Total number of possible arrangements for girls $= (3×2)$

Required probability

\(=\dfrac{(4\times 3\times 2\times 3\times 2)}{7!}\)

\(=\dfrac{1}{35}\)


(6) Comment(s)


Suman
 ()

IT can be like GBGBGBGBG like snehal said

then 5C3*3! for girls and 4! for boys


Ashok
 ()

The question specifically said that alternative position and choosing position other than BGBGBGB will result in at least 2 boys back to back which is not allowed. Now once this is understood, the 4 boys can shuffle themselves 4! and girls can 3!. so total favourable ways are 4!x3!

and prob is divide it by 7!


Snehal
 ()

Girls have 5 places to choose from ....at the start, between boys and a place after the last boy.....They can select the positions in 5C3 ways and then the girls can occupy the three places in 3! ways....therefore girls can be arranged in 5C3* 3! ways....Is my understanding right?



Pratik
 ()

What if que starts with a girl


VTG
 ()

then the boys will repeat like GBGBGBB which is not alternate,


Niranjana
 ()

It can also be 1/21