the probability of hitting a target is 3/4 and a man tried to hit this 5 times.what is the probability of hitting a target atleast thrice?

• using binomial theoram ncr q^(n-r)p^r
  n=5 p=3/4 q=1/4
  now atleast three so 5c3 1/4^(5-3)3/4^3 +5c4 1/4^(5-4)3/4^4 +5c5 1/4^(5-5)3/4^5
  on solving this equation getting answer
  answer is 459/512
• 10 years agoHelpfull: Yes(19) No(2)
• hitting target atleast 3 means 3times+4times+5 times.... so (3/4)^3*(1/4)^2+(3/4)^4(1/4)+(3/4)^5(1/4)^0 so ans is .1648
• 11 years agoHelpfull: Yes(12) No(8)
• THIS IS A BINOMIAL DISTRIBUTION PROBLEM
  probability of x successes in n trials is given asP(x)=nCx(p^x)q^(n-x)
  where x-no. of success
  n- number of attempts
  p- probability of success
  1-P(0)-P(1)-P(2)=1-.25^5-5*.75*.25^4-10*.75^2*.25^3=.896 IS THE ANSWER
• 12 years agoHelpfull: Yes(5) No(10)
• this is a question on binomial theorem of probability
  P(k success out of n trials)= nCk* p^k * q^(n-k)
  where
  p=probability of success
  q=probability of failure
  
  now in this question probability of success p=3/4 then probability of failur q=1-3/4=1/4
  therefore P(3 success out of 5 trials)=5C3* 3/4^3 * 1/4^2=135/512
  now it asks for at least 3 success then there may be 4 or 5 success also
  so, the required probability=P(3 success)+P(4 success)+P(5 success)
  P(at least 3 success)=.896
• 10 years agoHelpfull: Yes(5) No(1)
• The probability of hitting the target is 3/4.
  Therefore the probability of missing the target is 1-3/4=1/4.
  Hence the probability of hitting the target atleast thrice is 1-(1/4)^2.
  
• 11 years agoHelpfull: Yes(4) No(1)
• ) Let the probability of hitting the target be p = 3/4
  So probability of not hitting the target = q = 1 - 3/4 = 1/4
  No. of trials = n = 5
  Number of successes required = atleast 3; ==> r = 3, 4 & 5
  
  ii) By Binomial distribution probability of r successes in n trials is given by:
  P(x = r) = C(n, r)*{p^r}*{q^(n - r)}
  
  Applying the above, P(r = 3) = C(5,3)*{(3/4)^3}*{(1/4)^(2} = 270/(4^5)
  Similarly P(r = 4) = C(5,4)*{(3/4)^4}*(1/4) = 405/(4^5)
  P(r = 5) = C(5,5)*{(3/4)^5} = 243/4^5
  
  Thus total probability = (270 + 405 + 243)/(4^5) = 459/512
• 10 years agoHelpfull: Yes(3) No(1)
• ans)2880
  4 boys can occupy 4positions hence can be arranged in !4ways(_b1_b2_b3_b4_)
  4 girls can be arranged in !5ways because there are 5 positions for girls.
  so total no.of ways are= !4*!5= 24*120= 2880.
  
• 12 years agoHelpfull: Yes(2) No(7)
• i) Let the probability of hitting the target be p = 3/4
  So probability of not hitting the target = q = 1 - 3/4 = 1/4
  No. of trials = n = 5
  Number of successes required = atleast 3; ==> r = 3, 4 & 5
  
  ii) By Binomial distribution probability of r successes in n trials is given by:
  P(x = r) = C(n, r)*{p^r}*{q^(n - r)}
  
  Applying the above, P(r = 3) = C(5,3)*{(3/4)^3}*{(1/4)^(2} = 270/(4^5)
  Similarly P(r = 4) = C(5,4)*{(3/4)^4}*(1/4) = 405/(4^5)
  P(r = 5) = C(5,5)*{(3/4)^5} = 243/4^5
  
  Thus total probability = (270 + 405 + 243)/(4^5) = 459/512
  
• 9 years agoHelpfull: Yes(2) No(1)
• P(0 HIT)=1/256
  P(1 HIT)=15/256
  p(2 HIT)=90/256
  P(ATLEAST 3 HIT)=1- 1/256 - 15/256 -90/256= 75/128
• 12 years agoHelpfull: Yes(1) No(5)
• answer is -459/512
• 7 years agoHelpfull: Yes(0) No(0)
• Probability of hitting a target = Probability of success, p = 34
  
  Probability of not hitting = Probability of failure, q = 1−34=14
  
  Probability of hitting a target at least 3 times out of 5 times = sum of probabilities of hitting a target exactly 3 times, exactly 4 times and exactly 5times
  
  P(X≥3) = P(3) + P(4) + P(5)
  
  = 5C3(34)3(14)2+5C4(34)4(14)1+5C5(34)5
  
  = 10(2764)(116)+5(9256)(14)+1(2431024)
  
  = 2701024+451024+2431024
  
  = 5581024
  
  = 279512
  
  = 0.5449
• 6 years agoHelpfull: Yes(0) No(0)