Q. V Anand and Gary Kasparov play a series of 5 chess games. The probability that V Anand wins a game is 2/5 and the probability of Kasparov winning the game is 3/5.There is no probability of a draw.The series will be won by the person who wins 3 matches. Find the probability that Anand wins the series.(The series ends at the moment when any of the two wins 3 matches

Option
A)992/3125
B)273/625
C)1021/3125
D)1081/3125

• right ans:A
  as 3 cases r enlisted by frnds
  1)first 3 win by aanad=8/125
  2)4 matches played i.e aanad win 3 out of 4 except last 1 =3*8/125*3/5
  3)5 matches played i.e aanad win 3 out of 5 &in order to have all 5 matches aanad have to surely win lat 1 meaning by in earlier 4 matches he win only 2 =4c2*8/125*9/25
  by adding all three(8/125+72/625+432/3125) we have 992/3125
• 10 years agoHelpfull: Yes(4) No(0)
• it is solved by using binomial distribution
  probability of success=2/5
  probability of failure=3/5
  answer is 5c3(2/5)^3(3/5)^2
• 11 years agoHelpfull: Yes(0) No(1)
• @febin we are not getting the answer by using the method of binomial distribution and solving your equation
• 11 years agoHelpfull: Yes(0) No(0)
• Sumit...sry..here goes the correct solution...there are 3 cases.
  1)anand can win the 1st 3 games...3c3*(2/5)^3
  2)3 wins and 1 loss 0ut 0f four games....4c3*(2/5)^3*(3/5)
  3)3 wins and 2 loss out of 5......5c3*(2/5)^3*(3/5)^2
  Sum of these 3 cases vil give answer sumit
• 11 years agoHelpfull: Yes(0) No(0)