A man can hit a target once in 4 shots. If he fires 4 shots in succession GMAT Problem Solving

Question: 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) 1/256.
C) 81/256
D) 175/256
E) 144/256

“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?” - this is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. 

In the GMAT Problem Solving section, examiners measure how well the candidates make analytical and logical approaches to solve numerical problems. In this section, candidates have to evaluate and interpret data from a given graphical representation. In this section, mostly one finds out mathematical questions. Five answer choices are given for each GMAT Problem solving question. 

Solution and Explanation:

Approach Solution 1:

The first and the most simple approach to getting the answer has been explained below.

A man needs to ensure that he can hit the target with the help of 4 shots at least once. Accordingly, if one does not hit the target even once, then the 4 shots would be wasted and none of the shots could be used to hit the target.

Accordingly, if the man can hit the target once, then it might miss the rest of the 3 shots. For hitting, once out of four shots, the probability stands at ¼. However, if the other shots are missed then, the probability stands with ¾ shots of the target missed obtained with 1- ¼ = ¾

The Probability of not hitting the target in any of the 4 shots = (3÷4)* (3÷4)*(3÷4)= 81÷256

The Probability of hitting the target at least once using the 4 shots is equal to 1 - 81/256 = 175/256. Hence, the correct answer is C

Correct Answer: C

Approach Solution 2:
It is stated in the question that a man can hit the target if he hits it at least once. There is another method to solve the question which is given below.

Say, if you hit the target on the first shot, then what will be the outcomes of the remaining shots do not matter.

If you do not hit the target on the first shot, but hit it on the second shot, again, the man has hit the shot on the second time. So the rest of the two shots left does not matter.

If you do not hit the first and the second shot but hit with the third one, the remaining shot does not matter because once, one out of the four shots has been used.

If you do not hit the first, the second shot, and the third shot but hit the fourth one, then at least once the target will be hit. The remaining one shot does not matter because once, one out of the four shots has been used.

Significantly, the Probability of hitting the target = (1÷4)+(3÷4)*(1÷4)+(3*4)*(3÷4)*(1÷4)+(3*4)*(3÷4)*(3÷4)*(1÷4)

This equals to (1*64 + 3*16 + 9*4 + 27)/256 1*64+3*16+9*4+27÷256

Based on these equations that have been incorporated the method could be helpful in yielding the result of 175/256.

Correct Answer: C

Approach Solution 3:

The third method of finding out the answer for the probability of hitting the target once with four shots can be identified with the probability of each shot.

This implies that the probability of the first shot hitting the target and the rest of the three shots missing the target can be equated as:

(1÷4)*(3÷4)*(3÷4)*(3÷4)*4!÷3!=27*4÷256

Here, the above 4!/3! has been placed because it implies 4 shots out of which 1 will hit the target and 3 will be missed. Hence, 3 of the shots are identical.

The probability of the target hitting twice is equal to the:

(1÷4)*(1÷4)*(3÷4)*(3÷4)*4!÷2!*2!=54÷256

Here, the above 4!/3! has been placed because it implies 4 shots out of which 2 will hit the target and 2 will be missed. Hence, the 2 pairs of shots are identical.

The probability of the target hitting thrice is equal to the:

(1÷4)*(1÷4)*(1÷4)*(3÷4)*4!÷3!=12÷256

The probability of the target hitting four times is equal to the:

(1÷4)*(1÷4)*(1÷4)*(1÷4)*4!÷3!=1÷256

By adding all the values implying- \(\frac{27*4}{256}+\frac{54}{256}+\frac{12}{256}+\frac{1}{256}=\frac{175}{256}\)

Correct Answer: C

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