There are 5 different boxes named a b c d e there are two balls red and green a box

unable to understand the question
10 years agoHelpfull: Yes(11) No(3)
lets see how many blue balls can be adjacent to each other:
B,B,B,B,B -1
B,B,B,B------>B,B,B,B,R ; R,B,B,B,B -2
B,B,B-------->B,B,B,R,R ; R,B,B,B,R ; R,R,B,B,B -3
B,B---------->B,B,R,R/B,R ; B,B,R,R,R/B ; R,B,B,R,R/B ; R/B,R,B,B,R ; R,R/B,R,B,B ; R/B,R,R,B,B -6*2 =12
B,B,R,B,B - 1
ie total= 19
so total = 2^6-19 = 13
ie 13/32
10 years agoHelpfull: Yes(6) No(0)
how total outcome is 32??
10 years agoHelpfull: Yes(5) No(0)
as there are 5 boxes and 2 balls to be filled therefore total outcomes
2^5=32
10 years agoHelpfull: Yes(3) No(0)
Its 13/32..
There will be 13 such combinations

RBRBR , BRBRB , RRRRR , RRRRB and so on...
Ans will be 13/32
10 years agoHelpfull: Yes(2) No(0)
The denominator is 32 because it represent all the cases, which (according to p&c ) is 2*2*2*2*2.
Every box has 2 choices, to have a Blue ball or a red ball.
5 years agoHelpfull: Yes(1) No(0)
how denominator is 32?
10 years agoHelpfull: Yes(0) No(0)

There are 5 different boxes named a,b,c,d,e. there are two balls red and green.a box is to be filled with one ball either red or blue. what is the probability that no two blue balls are adjacent to each other.