Latest Elitmus Aptitude Question SOLUTION: there are red and blue balls which can be filled in 5 boxes. all balls are similar except color. what is the probability that no two consecutive boxes have blue balls.

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Probability

Elitmus Question Papers:

(#M40104623) ELITMUS Numerical Ability question probability Keep an EYE

there are red and blue balls which can be filled in 5 boxes. all balls are similar except color. what is the probability that no two consecutive boxes have blue balls.

Asked In Elitmus Ayushee Singh (9 years ago)

Unsolved Read Solution (8)

Is this Puzzle helpful?   (5)   (2) Submit Your Solution Probability

Read Solution (8) :

vipul kumar  (9 years ago)

Ans: 12
Exp/ A/q we hv sufficient no of R n B balls. We hv to focus only that no consecutive 2 balls having Blue ballls.
Let we hv chose only ...1 B and 4 R balls...., So possible combinations are-
(B,R,R,R,R), (R,B,R,R,R), (R,R,B,R,R), (R,R,R,B,R), (R,R,R,R,B). => 5ways
Now assume we hv choose ....2 B and 3 R balls.., So possible combinations are-
(B,R,B,R,R), (B,R,R,B,R), (B,R,R,R,B), (R,B,R,B,R), (R,B,R,R,B), (R,R,B,R,B) => 6ways
(simply we r placing Blue balls at places 1-3,1-4,1-5,2-4,2-5, 3-5)

similarly for 3 Blue balls, we hv only 1 option
(B, R, B, R, B)...........................................................................................=> 1 ways
we cant select 4 and 5 blue balls bcoz in these case we found 2 blue balls
at consecutive place!!
Total 5+6+1= 12 Ans.

Is this solution Helpfull? Yes (24) | No (4)  

amit singh  (9 years ago)

There are 5 spaces each can be filled by either blue or red so total no. of ways are2^5
And no. of favourable outcomes are 12 hence
12/2^5 or 3/8

Is this solution Helpfull? Yes (9) | No   

Lavika Kurda  (9 years ago)

The question demands probability..not the number of ways. Could anyone of you please put some light on that?

Is this solution Helpfull? Yes (6) | No   

Rahul Ranjan  (9 years ago)

13/32 is d ans

Is this solution Helpfull? Yes (5) | No   

sonu kumar  (9 years ago)

let us find what is the prob. that two consecutive boxes have blue color
5 boxes, TOTAL WAYS=2^5=32;
consecutive boxes have blue color, total way= 23(simple permutation )
so, prob. of consecutive boxes have blue color= 23/32;
required probability= 9/32

Is this solution Helpfull? Yes (4) | No   

Ravindra Pratap Singh  (9 years ago)

Possibilies 1.1 blue 4 red 2ND 2blue 3 red
3Rd 3blue 2 red.
For possibility 1St there are 5 ways as blue can be placed in any of 5 boxes
For possibility 3Rd we have. 1 way only BRBRB
For possibility 2ND
If ist place is blue BR_ _ _ that is 3 options for 2nd blue ball
If Ist place is red RBR_ _ that is two options for 2nd blue
If Ist and 2nd place both red only possibility is RRBRB
HENCE TOTAL NO OF WAYS =5+(3+2+1)+1=12

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

subhajit Roy   (9 years ago)

Total no of ways filling 5 boxs with either blue or red balls 2^5=32
Two adjacent boxes with blue 4 ways ie 12,23,34 and 45.
Three adjacent boxes with blue 3ways ie 123,234,345.
Four adjacent 1234,2345
And 5 adjacent 1 way
Hance total =(4+3+2+1)=10
The no of ways of filling up the boxes such dt no 2 adjacent boxes hv blue =32-10=22

Is this solution Helpfull? Yes (2) | No   

abhishek  (9 years ago)

ani is -3333

Is this solution Helpfull? Yes (1) | No   

Please Login to Read Solution.