A student has answered 7 of the 10 questions on the true-false test co

Question

A student has answered 7 of the 10 questions on the true-false test correctly and has decided to guess randomly on the rest. If all of the questions are equally weighted, what is the probability of the student receiving a score of 80% or more on the test?

A. 0.125
B. 0.5
C. 0.7
D. 0.8
E. 0.875

A. 0.125
B. 0.5
C. 0.7
D. 0.8
E. 0.875

sobby · Accepted Answer

out of 3 question the student have to correct 1 .
prob. of getting answer correct = 1/2
prob. of getting answer incorrect = 1/2 
let the student do all 3 answer incorrect = prob = 1/2 * 1/2 * 1/2
So getting at least one correct = 1-1/8 = 7/8 
P = 7/8 = E

mohshu · Answer

80% or more,,implies correct answers can be 8, 9 or 10..

three ways ,,YYY, YYN, YNN....
YYY = one way=1/8
YYN = 3 * 1/8
YNN = 3 *1/8

add it up == 0.875
ans E

matthewsmith_89 · Answer

mohshu  I do not understand your solution. Can you explain this in depth. Anyone else is free to explain
mohshu solution in depth as well. Thanks.

pushpitkc · Answer

YYN can have three combinations (YYN,NYY or YNY) (i.e) you have to get one wrong. It could either be the 8th question, the 9th question or the 10th question

Similarly you have three combinations for YNN(YNN,NYN or NNY). Hope it helps!

JDCal8899 · Answer

Instead of thinking about all the ways to get 80% or more, think about how not to get 80%.

If the student is already at 70%, the student only needs to get one correct to have 80%. Therefore, the only way not to pass is to get all three wrong.

(1/2) * (1/2) * (1/2) = 1/8, the odds of three incorrect

1 - 1/8 = 7/8, the odds of anything else, or getting 80% or higher.

7/8 = .875 choice E

JeffTargetTestPrep · Answer

In order for the student to get an 80% or higher on the test, he must answer at least 1 of the last 3 questions correctly. So, we can use the following formula:

1 = P(answering at least 1 of last 3 questions correctly) + P(answering none of the last 3 questions correctly)

The probability of answering none of the last 3 questions correctly is 1/2 x 1/2 x 1/2 = 1/8.

Thus, the probability of answering at least 1 of the last 3 questions correctly is 1 - 1/8 = 7/8 = 0.875.

Answer: E

plasticplastic · Answer

In order for the student to get at least 80% that means the student has to get at least one more correct on the remaining three questions.

Another way to think about it is that all the student has to avoid is getting all the remaining questions wrong. If we find the probability of getting all the questions wrong, then we will simultaneously know what the probability of NOT getting all of the questions wrong is by subtracting this by 1. This is just another way of thinking of how to find out how to get at least one question right.

The probability of getting a True or False question wrong is 1/2. 
The probability of getting a True or False question wrong on all three questions is 1/2 * 1/2 * 1/2 = 1/8. 
Therefore the probability of NOT getting all three questions wrong is 1 - 1/8 = 1 - 0.125 = 0.875.

Manishcfc8 · Answer

Just one more question needed to be correct to reach 80%.
P(at least one correct of remaining 3) = 1- P(no correct of the three)

P(no correct of the three) = P(1&2&3)= P(1).P(2).P(3) since the three are independent questions, all of which is equal to 0.5
Which means above probability is 0.125

P(at least one correct) = 1-0.125= 0.875

Sent from my DUK-L09 using  GMAT Club Forum mobile app

merajul · Answer

A student has answered 7 of the 10 questions on the true-false test correctly and has decided to guess randomly on the rest. If all of the questions are equally weighted, what is the probability of the student receiving a score of 80% or more on the test?

A. 0.125
B. 0.5
C. 0.7
D. 0.8
E. 0.875

A. 0.125
B. 0.5
C. 0.7
D. 0.8
E. 0.875

The student answered 7 out of 10 correct. The question asks for P(A) , where P(A) >= 80%

This means, student has to answer either 8th, 9th & 10th question or combined correctly.

No of questions =3 
Outcomes / question = 2 ( Right or Wrong)
So number of possibilities = 2 ^ 3 =8

Possibility of getting 80% = 3C1/8 = 3
Possibility of getting 90% = 3C2/8 = 3
Possibility of getting 100% = 3C3/8 = 1

So P(A) = (3C1 + 3C2 + 3C3) / 8 = 7/8 = 0.875 (Answer-E)

bumpbot · Answer

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A student has answered 7 of the 10 questions on the true-false test co

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