Probability of once in four

rajataga wrote:A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A. 1

B. 255/256

C. 175/256

D. 1/4

E. 1/2

There are 4 different scenarios in which he will "hit his target".

S1: hit once, miss 3 times
S2: hit twice, miss twice
S3: hit 3 times, miss once
S4: hit all 4 times

One way we could answer this question is to calculate the probability of each scenario and add them together.

However, a much quicker way to answer the question is to recognize that there's only 1 scenario we do NOT want: miss all 4 times. We can use our trusty "one minus" probability formula:

Prob(want) = 1 - Prob(don't want)

So, the probability of missing all 4 is:

3/4 * 3/4 * 3/4 * 3/4 = 81/256

Therefore, the probability of hitting at least one target is:

1 - 81/256 = 175/256... choose (C).

Thanks Stuart, you are correct.

However, what is the reason that in this scenario, you cannot think of them as 4 exclusive events, and say that if he takes 4 shots at the target, he will hit it atleast once?

i guess i know the answer mathematically....

each of these events are exclusive, and the outcome of one doesn't affect the outcome of another.....However, i am not yet convinced about this....

rajataga wrote:i guess i know the answer mathematically....

each of these events are exclusive, and the outcome of one doesn't affect the outcome of another.....However, i am not yet convinced about this....

Right.. the events are independent, which means the outcome of one doesn't affect the outcome of another.

Think about rolling a 6-sided die. There's a 1 in 6 chance of rolling each number. Does that mean that if you roll the die 6 times, you'll always get a 1, 2, 3, 4, 5 and 6? Of course not!

Of course, part of the problem is the wording of the question you posted, which is far too ambigious for an actual GMAT question. On the actual test, the question would likely have read something along the lines of:

"The probability of a man hitting a target with any shot he takes is 1/4. If the man fires 4 shots at the target, what's the chance that he hits it at least once?"

What's the source of the question?

Came across it on 4gmat.com....

Actually very true, the question is kinda worded wrong if u think about the first part...

After considering the scenarios 1 by 1 and adding them up, I still dont get it right. Could someone plse check my formula?

P(TMMM) + P(TTMM) + P(TTTM) + P(TTTT)

WHERE:
P(TMMM)= (1/4) * (3/4)^1/3 * 4(ways of hitting target once)
P(TTMM)= (1/4) ^1/2 * (3/4)^1/2 * 6
P(TTTM)= (1/4) ^1/3 * (3/4) * 4
P(TTTT)= (1/4) ^1/4 * 1

I am sure the problem lies in my "ways of hitting target" calculation.

Thanx in advance.