Probability
Binomial Distribution
Grade 12
Question:
<p>For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is \(\frac{4}{5}\), then the probability that he is unable to solve less than two problems is:</p>
<p>\(\dfrac{201}{5}\left(\dfrac{1}{5}\right)^{49}\)</p>
<p>\(\dfrac{316}{25}\left(\dfrac{4}{5}\right)^{48}\)</p>
<p>\(\dfrac{54}{5}\left(\dfrac{4}{5}\right)^{49}\)</p>
<p>\(\dfrac{164}{25}\left(\dfrac{1}{5}\right)^{48}\)</p>
Step-by-Step Solution
Key Concept: The candidate must solve either 49 or 50 problems (equivalently, fail at most 1 problem). Use binomial probability for exactly 0 and exactly 1 failures, then sum them.
<p><strong>Step 1: Interpret the condition</strong></p><p>'Unable to solve less than two problems' means the number of unsolved problems is 0 or 1 (at most 1 failure).</p><p><strong>Step 2: Define the distribution</strong></p><p>Let X = number of problems the candidate cannot solve.</p><p>Probability of not solving a problem: p = 1/5</p><p>Probability of solving a problem: q = 4/5</p><p>n = 50 problems</p><p><strong>Step 3: Calculate P(X ≤ 1)</strong></p><p>P(X ≤ 1) = P(X = 0) + P(X = 1)</p><p><strong>Step 4: Find P(X = 0)</strong></p><p>P(X = 0) = C(50,0) × (1/5)⁰ × (4/5)⁵⁰ = (4/5)⁵⁰</p><p><strong>Step 5: Find P(X = 1)</strong></p><p>P(X = 1) = C(50,1) × (1/5)¹ × (4/5)⁴⁹ = 50 × (1/5) × (4/5)⁴⁹</p><p>= 50 × (4/5)⁴⁹ × (1/5) = 10 × (4/5)⁴⁹</p><p><strong>Step 6: Add the probabilities</strong></p><p>P(X ≤ 1) = (4/5)⁵⁰ + 10(4/5)⁴⁹</p><p>= (4/5)⁴⁹[(4/5) + 10]</p><p>= (4/5)⁴⁹ × (54/5)</p><p>= 54(4/5)⁴⁹/5</p><p>∴ Answer: C</p>
Correct Answer: C