In a horse racing competition there were 18 numbered 1 to 18.the organisers assigned a probablity of winning the race to each horse such that horse1 would win is 1/7,for horse 2 it is 1/8 and for horse 3 it is 1/8.Assuming that tie is not possible.find chance that one of these three will win the race???
ans-23/56

• p(AUBUC)=P(A)+P(B)+P(C)-(P(A^B)+P(A^C)+P(B^C))+P(A^B^C)
  = 1/7+1/8+1/8-1/56-1/64-1/56+1/392
  =154/392
  =22/56
• 9 years agoHelpfull: Yes(10) No(5)
• as there cannot be a tie, the probability is 1/7+1/8+1/8=22/56
• 9 years agoHelpfull: Yes(10) No(4)
• Given 3 events are mutually exclusive so
  p(AUBUC)=P(A)+P(B)+P(C)-(P(A^B)+P(A^C)+P(B^C))+P(A^B^C)
  =1/7+1/8+1/8-0-0-0+0 (bcaz events are mutually exclusive so intersections are 0)
  =22/56
  =11/28
• 9 years agoHelpfull: Yes(6) No(1)
• answer is 19/64
  
  
• 9 years agoHelpfull: Yes(6) No(0)
• The chances of Horse No 1 winning 1/7 Probability of horse No 1 not winning 6/7
  
  The chances of horse No 2 winning 1/8 Probability of horse No 2 not winning 7/8
  
  The chances of horse No 3 winning 1/8 Probability of horse No 3 not winning 7/8.
  
  The chances of any one of these three horses winning the race will be:
  
  (A winning, B and C not winning) + (B winning and C and A not winning) + (C winning and A and B not winning)
  
  Now substituting the relevant information in the above we get the following.
  
  Probability of any one of the three horses winning is arrived at as
  
  (1/7 + 7/8 + 7/8) + ( 1/8 + 6/7 + 7/8) + ( 1/8 + 6/7 + 7/8) Reducing this we get the value
  
  Of probability of any one of these three horses winning as 19/64.
• 9 years agoHelpfull: Yes(4) No(1)
• Probability of Horse 1, 2 and 3 winning is 1/7, 1/8, 1/8 respectively.
  Probability of Horse 1, 2 and 3 not winning is (1-1/7), (1-1/8), (1-1/8) i.e., 6/7, 7/8,7/8 respectively.
  There are three possible situations which will satisfy the given conditions:
  a) Horse 1 win and horses 2 and 3 lose. Probability of this event happening = (1/7*7/8*7/8) = 7/64
  b) Horse 2 win and horses 1 and 3 lose. Probability of this event happening = (6/7*1/8*7/8) = 6/64
  c) Horse 3 win and horses 1 and 2 lose. Probability of this event happening = (6/7*7/8*1/8) = 6/64
  Hence, the required probability is = (7/64 + 6/64 + 6/64) = 19/64.
• 8 years agoHelpfull: Yes(3) No(0)
• Required prob= (1/7)(1-1/8)(1-1/8)+(1/8)(1-1/7)(1-1/8)+(1/8)(1-1/7)(1-1/8)
  =(1/7)(7/8)(7/8)+(1/8)(6/7)(7/8)+(1/8)(6/7)(7/8)
  =61/64
• 9 years agoHelpfull: Yes(2) No(2)
• Probability of Horse 1, 2 and 3 winning is 1/7, 1/8, 1/8 respectively.
  Probability of Horse 1, 2 and 3 not winning is (1-1/7), (1-1/8), (1-1/8) i.e., 6/7, 7/8,7/8 respectively.
  There are three possible situations which will satisfy the given conditions:
  a) Horse 1 win and horses 2 and 3 lose. Probability of this event happening = (1/7*7/8*7/8) = 7/64
  b) Horse 2 win and horses 1 and 3 lose. Probability of this event happening = (6/7*1/8*7/8) = 6/64
  c) Horse 3 win and horses 1 and 2 lose. Probability of this event happening = (6/7*7/8*1/8) = 6/64
  Hence, the required probability is = (7/64 + 6/64 + 6/64) = 19/64
• 8 years agoHelpfull: Yes(2) No(0)
• 1/7+1/8+1/8 =21/56
• 9 years agoHelpfull: Yes(1) No(2)