Latest Elitmus Aptitude Question SOLUTION: There are 4 boys and 3 girls.they sit in arrow randomly.what is the probability that all girls are together? A. 1/14 B. 2/14 C. 5/14 D. 3/14

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(#M40165041) ELITMUS Numerical Ability question Keep an EYE

There are 4 boys and 3 girls.they sit in arrow randomly.what is the probability that all girls are
together?

A. 1/14 B. 2/14 C. 5/14 D. 3/14

Asked In Elitmus OMKAR GANDHI (6 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

vijay pandey  (6 years ago)

Let us assume the 3 girls as a single unit.

So we've now 4 boys +1 = 5 units which can arrange themselves in 5! = 120 ways. Now in these , 3 girls can arrange themselves in 3!= 6 ways.

So finally we've the favourable cases= 120*6=720

The sample space consists of all possible arrangements of 4 boys and 3 girls which is = (4+3)!= 7! = 5040

Required probability = (favourable cases)/(sample-space)=720/5040 =72/504= 1/7=2/14

Is this solution Helpfull? Yes (31) | No   

Javlak Mansuri  (6 years ago)

4 boys and 3 girls total 7 people
So total way of arrangement these 7 people=7!
so n(s)=7!
now let us consider all three girls as single one.
now 4 boys-1 girl(because we considered 3girls as 1) total 5 people
So number of way of arrangement=5!
again group of girls that we considered as 1 can be arrange by 3! so total favourable ways n(e)=5!*3!
required probability=5!*3!/7!
after solving ans would be = 1/7

Is this solution Helpfull? Yes (4) | No   

pallabhi swain  (6 years ago)

b.2/14

the probability that 3 girls can sit together is 3!*5! (___1st boy___2nd boy___3rd boy___4th boy___)and the total arrangement is 7! so 3!*5!/7!=2/14

Is this solution Helpfull? Yes (2) | No (5)  

shanu mittal  (6 years ago)

Simple!!

Taking 3 girls as a single unit (coz they sit together) total students are now 5 & the arrangements foe 5 will be 5!
Now, 3 girls will have their own arrangements which will be 3!.
Total students are 7.
So, (3! * 5!) / 7! = 2/14

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Syed Arshi  (4 years ago)

2/14 option B is correct

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