Question

Suppose the probability of a surprise test on any given day is 50 per cent. What is the probability that there are exactly 2 surprise tests in a 6-day period?

• A. \( \frac{5}{32} \)
• B. \( \frac{3}{16} \)
• C. \( \frac{7}{32} \)
• D. \( \frac{15}{64} \)
• E. \( \frac{9}{32} \)

Hint:

Recognize the keywords for a binomial probability problem: a fixed number of trials, independent trials, only two outcomes per trial, and a constant probability of success. When \(p=q=1/2\), the formula simplifies to \( C(n,k) / 2^n \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a problem of binomial probability. We have a fixed number of independent trials (6 days), each with two possible outcomes (test or no test), and the probability of success is constant for each trial.
Step 2: Key Formula or Approach:
The binomial probability formula is: \[ P(X=k) = C(n, k) \cdot p^k \cdot q^{n-k} \] Where:

\(n\) is the total number of trials.
\(k\) is the exact number of successful outcomes.
\(p\) is the probability of success on a single trial.
\(q\) is the probability of failure on a single trial (\(q = 1-p\)).
\(C(n, k)\) is the number of combinations, calculated as \( \frac{n!}{k!(n-k)!} \).
Step 3: Detailed Explanation:
From the problem statement:

Number of trials, \(n = 6\) (a 6-day period).
Exact number of successes, \(k = 2\) (exactly 2 surprise tests).
Probability of success (a test), \(p = 50% = 0.5 = \frac{1}{2}\).
Probability of failure (no test), \(q = 1 - 0.5 = 0.5 = \frac{1}{2}\).
First, calculate the number of ways to choose 2 days out of 6 for the tests, \(C(6, 2)\): \[ C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15 \] There are 15 different combinations of 2 test days in a 6-day period.
Next, calculate the probability for any one of these specific combinations. For example, tests on the first two days and no tests on the next four (TTNNNN): \[ P(\text{one combination}) = p^k \cdot q^{n-k} = \left(\frac{1}{2}\right)^2 \cdot \left(\frac{1}{2}\right)^{4} = \left(\frac{1}{2}\right)^6 = \frac{1}{64} \] Finally, multiply the number of combinations by the probability of one combination: \[ P(\text{exactly 2 tests}) = C(6, 2) \times P(\text{one combination}) = 15 \times \frac{1}{64} = \frac{15}{64} \] Step 4: Final Answer:
The probability of having exactly 2 surprise tests in a 6-day period is \( \frac{15}{64} \).