the probability of a bomb hitting a bridge is 1/2 and twodirect hits are neede to destroy it.the least no of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is?

• P(bomb hit)=1/2=P(bomb doesn't hit)
  atleast 2 bombs are required mean we have to calculate the probability to hit atleast 2 bombs out of n(suppose) bombs.So, this can be done like
  P(atleast 2 bombs hit)=1-P(zero bomb hit)-P(1 bomb hit)
  =1-all failure-one success other failure
  =1-[(1/2)*(1/2)*(1/2)....n]-one hit*(n-1)fail*n ways to arrange hit with
  failures
  =1-(1/2)^n-n*(1/2)*[(1/2)^(n-1)]
  now according to given condition
  0.9≥ P(atleast two hit) i.e 1-(1/2)^n-n*(1/2)*[(1/2)^(n-1)]
  on solving this we have for n=6 optimum value.
• 11 years agoHelpfull: Yes(4) No(6)
• Since the probability of the bomb hitting and missing the bridge is equivalent, we can consider the situation to be like a coin toss (Heads and Tails). So for example if ten bombs are dropped, the probability of ten hits is equal to the probability of ten missed. If we consider the cases in binary (1 for a hit and 0 for a miss), we have the following:
  For 2 bombs, the total combinations are 4 (2^2).
  Miss Miss = 0 0
  Miss Hit = 0 1
  Hit Miss = 1 0
  Hit Hit = 1 1
  
  So the probability of atleast two hits = 1-(Probability of 1 hit + Probability of no hits)
  In binary, Probability of 1 hit + Probability of no hits would be the number of cases where we have a single one and the number of cases we have all zeros. This is divided by the total possible number of cases and subtracted from one to get the probability.
  For n = 5, Probability = 1 - ( (5+1)/32 ) = 0.8125
  For n = 6, Probability = 1 - ( (6+1)/64 ) = 0.8906
  For n = 7, Probability = 1 - ( (7+1)/128 ) = 0.9375
  
  Hence the number of bombs to be dropped is 7.
• 7 years agoHelpfull: Yes(3) No(1)
• 3.735365e-10
• 11 years agoHelpfull: Yes(1) No(2)