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Given

    • A math exam has five true or false questions.
    • Alan selects one answer for every question (Assuming it is done randomly).

To Find

    • The probability of getting at least 3 answers correct.


Approach and Working Out


    • Assuming he is choosing it randomly.
    • It can be done applying C(n,r) = C(n, n-r) concept.
      o 3 Correct, 4 Correct, or 5 correct will have the same probability as getting 2 correct, 1 correct and 0 correct respectively.
      o So the answer has to be ½.

Correct Answer: Option E
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Quote:
A math exam has five true or false questions. If Alan selects one answer for every question, what is the probability of getting at-least 3 answers correct?
Step 1: Understanding the question
As there are five questions, out of the total possible ways of answering the question, half of the ways number of correct answers will be more than the number of incorrect answers, and the remaining half of the ways number of incorrect answers will be more than the number of correct answers.

Step 2: Calculation
In other words, at-least 3 correct answers out of 5 questions means more number of correct answers than number of incorrect answers, = 1/2

E is correct
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QuantMadeEasy
Q62. A math exam has five true or false questions. If Alan selects one answer for every question, what is the probability of getting at-least 3 answers correct?

a. 1/6

b. 1/5

c. ¼

d. 1/3

e. ½

The following cases are possible:

WWRRR
\(\frac{5!}{2!3!} =\) 10 cases

WRRRR
\(\frac{5!}{4!}\) = 5 cases

RRRRR
\(\frac{5!}{5!}\) = 1 case

Hence total favorable cases = 10+15+1 =16

Overall total cases 2*2*2*2*2 = 32

Hence probability = \(\frac{16}{32} =\frac{1}{2} \)

Ans- E

Hope it's clear.
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