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.

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(#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)

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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.

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amit singh  (8 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?

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Rahul Ranjan  (8 years ago)

13/32 is d ans

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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

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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

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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

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abhishek  (9 years ago)

ani is -3333

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