cicerone - can you explain how you came up with that anser. I went about this problem with trying to figure out the probability that he will not hit the target twice. Thanks
Prob of hitting atleast twice = 1- (prob of hitting 0 times + prob of hitting 1 time)
Prob of hitting 0 times = 3/4x3/4x3/4x3/4 = 81/256.
Prob of hitting 1 time out of 4 times:
We have 4 cases here i.e. he could hit once in anyone of the 4 hits.
i.e if he hits in the first shot the prob is 1/4x3/4x3/4x3/4 =27/256
if he hits in the second shot the prob is 3/4x1/4x3/4x3/4 =27/256
Continuing this way it would be 4x27/256 = 108/256.
So 1-81/256-108/256 = 67/256.
The other way is
prob of hitting 2 times + prob of hitting 3 times + prob of hitting 4 times.
Again hitting 2 out of 4 we will have 6 cases.
For each case the prob will be 1/4x1/4x3/4x3/4 = 9/256
So prob of hitting 2 out of 4 is 6x9/256 = 54/256
Prob of hitting 3 out of 4 we will have 4 cases
For each case the prob will be 1/4x1/4x1/4x3/4 = 3/256
So prob of hitting 3 out of 4 is 4x3/256=12/256
Prob of hitting 4 out of 4 we will have only 1 case
So prob of hitting 4 out of 4 is 1/4x1/4x1/4x1/4 = 1/256.
SO prob of hitting atleast twice = 54/256+12/256+1/256 = 67/256
So either way it is 67/256.
But i think the first way is faster
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)