Matrix02 wrote:

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

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