if there a 5 chips ,3 are red and 2 are white.
and they are drawn randomly one after another.
what is the probability that the last chip is white
dont remember the options

• total ways= 5!/3!*2!=10
  ideal ways>> fix the white at last nd we r left with 3 red nd 1 white = 4!/3!=4
  hence probability= 4/10=2/5
• 9 years agoHelpfull: Yes(51) No(0)
• fix last chip as white.
  there will be 4 favorable ways
  RRRWW or RRWRW or RWRRW or WRRRW
  (3/5*2/4*1/3*2/2*1)+(3/5*2/4*2/3*1/2*1)+(3/5*2/4*2/3*1/2*1)+(2/5*3/4*2/3*1/2*1)
  =48/120
  =2/5
  
• 9 years agoHelpfull: Yes(37) No(2)
• condition is the last chip must be white there are two possibilities
  1. 1st one is red AND last one is white. OR
  2. 1st one is white AND last one is also white .
  3/5*2/4+2/5*1/5 = 2/5 ans
• 9 years agoHelpfull: Yes(1) No(0)
• Ans =2/5
  arrangements of n thing in a row in which r1 of same kind and r2 of another kind=n!/(r1!*r2!)
  
  total ways= 5!/3!*2!=10
  favourable ways>> fix the white at last & we r left with 3 red nd 1 white = 4!/3!=4
  hence probability= 4/10=2/5
• 8 years agoHelpfull: Yes(1) No(1)