me test series

Question

The probability of a man hitting a target in one fire is 3/4  The number of times at least he must fire at the target in order that his chance of hitting the target at least once will exceed  2/3 will be _______.

...........just help to solve for n

(3/4)^n < (1/3)

whether <3.85 or n>3.85(it is given as answer but how)

Comments

can you give the exact question it would be easy to understand!!!

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misprinted values are added please check it

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i think it'll always be more than 75%,

let's suppose he hit the target two times, then probability of not hitting the target both the times will be P(A`)=1/4*1/4, now hitting the target at least once will be 1-P(A`) = 15/16 ~ .9375.

if he hit the target three times, then probability of not hitting the target both the times will be P(A`)=1/4*1/4*1/4, now hitting the target at least once will be 1-P(A`) =63/64 ~ 0.984375.

Do they give solution or hint?

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and what you've asked to solve in words is, "number of time he should hit so that probability of hitting the target every time, less than(edit) 1/3."

which is surprisingly not "2/3" or "at least once".