The PROBABILITY of a shooter hitting a target is 3/4. How many minimum no. of times must he/she fire so that the probability of hitting the target atleast once is more than 0.99????? 

this can be done by by bernoulis trails.

p=probability of hitting the targt=3/4 and q=prob of not hitting the target=1/4

now probability of hitting targt at least once >0.99

• i.e. p(at least once)=1-p(x=0)

  = 1- cⁿ₀1/4ⁿ> 0.99

  cⁿ₀1/4n <0.01 i.e. 1/4ⁿ   <0.01 since cⁿ₀ =1

  i.e. 4ⁿ > 1/0.01=100

thus minimum value  to satisfy the inequality is 4 times{ans}