Probability
Binomial Distribution
Grade 12
Question:
<p>A candidate attempts 50 problems. The probability of solving any problem is \(\dfrac{4}{5}\). What is the probability that the candidate is unable to solve less than two problems (i.e., is able to solve either 50 or 49 problems)?</p>
<p>(1) \(\left(\dfrac{4}{5}\right)^{50} + 10\left(\dfrac{4}{5}\right)^{49}\)</p>
<p>(2) \(\left(\dfrac{4}{5}\right)^{49}\left[\dfrac{4}{5}+10\right]\)</p>
<p>(3) \(\dfrac{54}{5}\left(\dfrac{4}{5}\right)^{49}\)</p>
<p>(4) All of the above</p>
Step-by-Step Solution
Key Concept: This is a binomial probability problem where we need P(X ≥ 49) = P(X = 49) + P(X = 50) with n=50, p=4/5, q=1/5. The phrase 'unable to solve less than two' means 'fails on at most 1 problem' or 'solves at least 49 problems'.
<p><strong>Step 1:</strong> Identify the setup. Let X = number of problems solved. We have n=50, p=4/5 (probability of solving), q=1/5 (probability of not solving). 'Unable to solve less than two' means the candidate fails on fewer than 2 problems, i.e., solves at least 49 problems.</p><p><strong>Step 2:</strong> Calculate P(X=50): P(X=50) = C(50,50)(4/5)^{50}(1/5)^0 = (4/5)^{50}</p><p><strong>Step 3:</strong> Calculate P(X=49): P(X=49) = C(50,49)(4/5)^{49}(1/5)^1 = 50 · (4/5)^{49} · (1/5) = 50 · (4/5)^{49} · (1/5)</p><p><strong>Step 4:</strong> Add the probabilities: P(X ≥ 49) = (4/5)^{50} + 50 · (4/5)^{49} · (1/5) = (4/5)^{49}[(4/5) + 50·(1/5)] = (4/5)^{49}[(4/5) + 10] = (4/5)^{49}[(4+50)/5] = (4/5)^{49} · (54/5) = 54·(4/5)^{49} · (1/5) = 54·(4)^{49}/(5)^{50}</p><p>∴ Answer: D</p>
Correct Answer: D