probability that guy1 will succesfully hit a shot is 2/3. probabilty of guy2 succesfully hitting a shot is 2/3. if they shoot two shots each, what is the probability that they have equal number of hits?

• Consider these cases:
  for 0 hit each: (1 way)
  (1/3*1/3)[guy1]*(1/3*1/3)[guy2]
  for 1 hit each: (4 ways)
  (2/3*1/3)*(2/3*1/3), (2/3*1/3)*(1/3*2/3), (1/3*2/3)*(2/3*1/3), (1/3*2/3)*(1/3*2/3)
  for 2 hits each:(1 way)
  (2/3*2/3)*(2/3*2/3)
  
  Add all these cases, you will get answer as 11/27.
• 8 years agoHelpfull: Yes(37) No(13)
• why are we adding case 0 and case 1 ? the question asks for only 2 shots ..
• 8 years agoHelpfull: Yes(8) No(3)
• using binomial expansion nCr*(P)^r*(1-P)^(n-r), where r is the no of success and n is the attempts(shots taken).
  now as each are having equal hits we need to consider all the possibilities & there can be 3 such cases
  1) guy1 hits 2 out of 2 AND(*) guy2 also hits 2 out of 2.......so using the above formula we get 16/81
  OR(+)
  2) guy1 AND guy2 hit 1 out of 2.......we get 16/81
  OR
  3) both guy1 and guy2 hit 0 in 2 attempts......we get 1/81
  
  
  so adding 16/81 + 16/81 + 1/81 = 33/81 = 11/27
  
• 8 years agoHelpfull: Yes(7) No(3)
• In four shots he can hit once,twice,thrice,all hit the probability of hitting the target isp(1hit out of 4)+P(2hit out of 4)+p(3hit out of 4)+p(All hit) it is total probability-probability of not hitting yhr target =>1-(3/4*3/4*3/4*3/4) =>175/256
• 7 years agoHelpfull: Yes(0) No(1)