If Joe does a test with 3 questions and each question has answers A, B

Question

If Joe does a test with 3 questions and each question has answers A, B or C, then what is the probability he gets at least 2 questions correct by answering randomly?

a)  \frac{12}{27}

b)  \frac{5}{54}

c)  \frac{1}{4}

d)  \frac{7}{27}

e)  \frac{3}{8}

EthanTheTutor · Answer

The main source of mistakes in this problem is not taking into account the fact that these questions have 3 possible answer choices. Another thing to notice here is that the problem uses the phrase “at least” … which means we should consider finding the probability of obtaining the opposite of what we want, then subtracting that result from 1. But in this case, we see that the two things are symmetric. What we want involves as much calculation as the opposite of what we want.

In other words, we want: 2 right and 1 wrong, or 3 right and 0 wrong. The opposite of this would be: 1 right and 2 wrong, or 0 right and 3 wrong. With the same number of situations in either case, it should be easier to calculate directly what we want, instead of subtracting from 1.

Method 1: Sequential Events

First, name the events. If "R" represents "right," and "W" represents "wrong," then as described above, we want: either  RRW  or  RRR .

Then, count the arrangements:  RRW  can be arranged in 3 ways,  RRR  in 1 way.

Now, calculate the probability of each sequence of events:  RRW=\frac{1}{3}×\frac{1}{3}×\frac{2}{3} , and  RRR=\frac{1}{3}×\frac{1}{3}×\frac{1}{3}

Then, combine:  3×RRW+RRR=3×\frac{2}{27}+\frac{1}{27}=\frac{7}{27}

Answer D .

Method 2: Simultaneous Events

First, name what we want, and the total:  \frac{#\ of\ ways\ to\ get\ 2\ right\ (and\ 1\ wrong),\ or\ 3\ right}{#\ of\ ways\ to\ answer\ 3\ questions}

Don’t forget to include the fact that we have 3 answer choices for each question in the denominator!

# of ways to answer 3 questions  =\binom{3}{1}×\binom{3}{1}×\binom{3}{1}=27

# of ways to get 2 right (and 1 wrong)  =\binom{3}{2}×\binom{2}{1}=6  (first choose which 2 questions to get right, then choose which of the wrong answers to select in the question you’re getting wrong)

# of ways to get 3 right  =1×1×1=1  … yeah, that’s it! You only have 1 way to get each question right.

This gives:  \frac{#\ of\ ways\ to\ get\ 2\ right\ (and\ 1\ wrong),\ or\ 3\ right}{#\ of\ ways\ to\ answer\ 3\ questions}=\frac{6+1}{27}=\frac{7}{27}

Answer D .

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If Joe does a test with 3 questions and each question has answers A, B

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