A can hit a target 6 times in 7 shots. B can hit the target 4 times in 5 shots, C can hit 3 times in 4 shots. What is the chance that the target is damaged by exactly 2 shots?
a)18/140
b)30/140
c)40/140
d)50/140
e)54/140

• I think each person A ,B and C are taking one chance each and we are supposed to find prob of hitting target by exactly two shots out of 3 shots.
  so desired combinations are
  A...B....C
  Hit hit miss
  Hit miss hit
  Miss hit hit
  
  person A will hit the target six out of seven times, so his hit probability is 6/7, and his miss probability is 1/7.
  
  B's hit probability is 4/5, and his miss probability is 1/5.
  
  C's hit probability is 3/4, and his miss probability is 1/4.
  
  So the probability that the target will be in (Miss hit hit)condition after the shooting = (1/7) * (4/5) * (3/4) = 12/140.
  
  prob of condition (hit miss hit) (6/7) * (1/5) * (3/4) = 18/140.
  
  prob of condition (hit hit miss)= (6/7) * (4/5) * (1/4) = 24/140.
  
  The sum of the three probabilities = prob of hitting target exactly twice 54/140 ........ option e)
• 12 years agoHelpfull: Yes(25) No(1)
• (6/7)x(4/5)x(1/4)+(6/7)x(1/5)x(3/4)+(1/7)x(4/5)x(3/4) = 54/140
• 8 years agoHelpfull: Yes(2) No(0)
• (6/7*4/7*1/4)+(1/7*4/5*3/4)+(6/7*3/4*1/5) = 54/140
• 8 years agoHelpfull: Yes(1) No(0)
• the answer is 18/140
• 7 years agoHelpfull: Yes(1) No(2)
• (ab)/100/(ba)/100=1
• 7 years agoHelpfull: Yes(0) No(2)