Q 2 boxes 32 black n 31 red balls the probability of getting black ball is maximum the maximum probability

to maximize the probability we can send all but 1 black balls to other box making probability

1/2 (ie chosing one box) * 1(no. of black balls in it) + 1/2*31/62 (as we have no 31 black and 31 red balls in it)

finally we have 1/2*1 + 1/2*1/2 = 3/4
11 years agoHelpfull: Yes(28) No(3)

The ans is 0.75.
we will keep one black ball in one box so that the probability of getting that is=1/2*1=1/2
Now rest of 31 black ball will be placed in the other box so that to get the maximum probability and it is=1/2*31/62
So the maximum prob of getting black ball is=1/2+1/2*31/62=0.75
11 years agoHelpfull: Yes(19) No(0)
1/2+1/2*31/62=3/4
11 years agoHelpfull: Yes(5) No(5)
probability=(1/2)*(1/31)=1/62.
11 years agoHelpfull: Yes(4) No(22)
(1/2)*(0/31)+(1/2)*(32/32)=0+(1/2)=1/2
11 years agoHelpfull: Yes(4) No(10)
3/4 is the correct ans
9 years agoHelpfull: Yes(0) No(2)

Q. 2 boxes, 32 black n 31 red balls the probability of getting black ball is maximum the maximum probability is