A box Contain 3 Red and 2 White balls.the balls are taken out without replacement what is the probability that one of the White ball is always taken out at the End??

• if any of the white ball can come last then
  any of the three red balls and one of the white balls should be taken out first
  so..4/5*3/4*2/3*1/2=1/5
• 10 years agoHelpfull: Yes(21) No(1)
• if any of the white ball can come last then
  any of the three red balls and one of the white balls should be taken out first
  i,e. (4/5)*(3/4)*(2/3)=2/5 Ans.
  
• 10 years agoHelpfull: Yes(17) No(6)
• for 1st ball probability ( including one white ) = 4/5
  then it is done without replacement..so for 2nd withdrawal = 3/4
  then for 3rd..= 2/3.
  hence answer = 4/5*3/4*2/3 = 2/5
• 10 years agoHelpfull: Yes(9) No(3)
• ns-4/10 i.e 2/5
  total possibilities=5!/(3!*2!)=10
  possibility that white is last = 4!/(3!*1!)=4
  so,probability=4/10=2/5
• 7 years agoHelpfull: Yes(0) No(0)