Probability
Conditional Probability
Grade 12

Question:

<p>In a multiple choice question there are four alternative answers of which one or more than one is correct. A candidate will get marks on the question only if he ticks the correct answer. The candidate decides to tick answers at random. If he is allowed up to three chances of answer the question, then the probability that he will get marks on it is</p>
<p>(a) \(\frac{1}{3}\)</p>
<p>(b) \(\frac{2}{3}\)</p>
<p>(c) \(\frac{1}{5}\)</p>
<p>(d) \(\frac{2}{15}\)</p>

Step-by-Step Solution

Key Concept: Model the problem as selecting from possible correct answer combinations; probability is 1/(total combinations) for each attempt.
<p><strong>Step 1:</strong> There are $2^4 - 1 = 15$ possible non-empty subsets of 4 answers (possible correct answer combinations).</p><p><strong>Step 2:</strong> The candidate randomly selects from these 15 possibilities up to 3 times without repetition.</p><p><strong>Step 3:</strong> Probability of getting correct answer on first try = $\frac{1}{15}$.</p><p><strong>Step 4:</strong> Probability of getting correct on second try = $\frac{14}{15} \cdot \frac{1}{14} = \frac{1}{15}$.</p><p><strong>Step 5:</strong> Probability of getting correct on third try = $\frac{14}{15} \cdot \frac{13}{14} \cdot \frac{1}{13} = \frac{1}{15}$.</p><p><strong>Step 6:</strong> Total probability = $\frac{1}{15} + \frac{1}{15} + \frac{1}{15} = \frac{3}{15} = \frac{1}{5}$.</p><p>∴ Answer is (c).</p>
Correct Answer: B

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