Understanding the Probability Problem

This problem asks for the probability of a candidate solving a certain number of questions in an examination. We are given the total number of questions and the probability of solving a single question. This scenario fits the criteria for a binomial probability distribution.

Identifying the Binomial Distribution Parameters

A binomial distribution applies when there are a fixed number of independent trials, each with only two possible outcomes (success or failure), and the probability of success is constant for every trial.

• Number of trials (n): The total number of questions in the examination, which is 5.
• Probability of success (p): The probability that the candidate solves a single question, given as 1/2.
• Probability of failure (q): The probability that the candidate does not solve a single question. This is \(q = 1 - p = 1 - 1/2 = 1/2\).

Let \(X\) be the random variable representing the number of questions the candidate solves out of the 5. \(X\) follows a binomial distribution with parameters \(n=5\) and \(p=1/2\).

Calculating the Probability of Solving At Least 2 Questions

We need to find the probability that the candidate solves at least 2 questions. This means the number of solved questions \(X\) is greater than or equal to 2, i.e., \(P(X \ge 2)\).

\(P(X \ge 2)\) includes the probabilities \(P(X=2)\), \(P(X=3)\), \(P(X=4)\), and \(P(X=5)\). Calculating each of these and summing them up can be tedious.

An easier approach is to use the complement rule. The total probability is 1. \(P(X \ge 2)\) is equal to \(1\) minus the probability of solving less than 2 questions. Less than 2 questions means solving 0 questions or solving 1 question.

So, \(P(X \ge 2) = 1 - P(X < 2) = 1 - [P(X=0) + P(X=1)]\).

Binomial Probability Formula

The probability of getting exactly \(k\) successes in \(n\) trials is given by the binomial probability formula:

\(P(X=k) = \binom{n}{k} p^k q^{n-k}\)

where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient.

Calculating P(X=0) and P(X=1)

Let's calculate \(P(X=0)\), the probability of solving 0 questions:

\(P(X=0) = \binom{5}{0} (1/2)^0 (1/2)^{5-0}\)

\(P(X=0) = 1 \times 1 \times (1/2)^5\) (Since \(\binom{5}{0}=1\) and \((1/2)^0=1\))

\(P(X=0) = (1/2)^5 = \frac{1}{2^5} = \frac{1}{32}\)

Now, let's calculate \(P(X=1)\), the probability of solving 1 question:

\(P(X=1) = \binom{5}{1} (1/2)^1 (1/2)^{5-1}\)

\(P(X=1) = 5 \times (1/2)^1 \times (1/2)^4\) (Since \(\binom{5}{1}=5\))

\(P(X=1) = 5 \times (1/2)^{1+4} = 5 \times (1/2)^5\)

\(P(X=1) = 5 \times \frac{1}{32} = \frac{5}{32}\)

Calculating P(X < 2)

The probability of solving less than 2 questions is the sum of \(P(X=0)\) and \(P(X=1)\):

\(P(X < 2) = P(X=0) + P(X=1) = \frac{1}{32} + \frac{5}{32} = \frac{1+5}{32} = \frac{6}{32}\)

This fraction can be simplified by dividing the numerator and denominator by 2:

\(\frac{6}{32} = \frac{3}{16}\)

Calculating P(X ≥ 2)

Finally, we can find the probability of solving at least 2 questions:

\(P(X \ge 2) = 1 - P(X < 2) = 1 - \frac{3}{16}\)

To subtract, we express 1 as a fraction with denominator 16:

\(1 = \frac{16}{16}\)

So,

\(P(X \ge 2) = \frac{16}{16} - \frac{3}{16} = \frac{16-3}{16} = \frac{13}{16}\)

The probability that the candidate was able to solve at least 2 questions is 13/16.

Summary of Probability Calculations

Revision Table: Probability Concepts

Additional Information: Understanding "At Least" Probability

In probability questions, the phrase "at least \(k\)" means the number of successes is \(k\) or more (\(X \ge k\)). Similarly, "at most \(k\)" means the number of successes is \(k\) or less (\(X \le k\)). Understanding these phrases is crucial for setting up the problem correctly.

• "At least 2" means \(X = 2, 3, 4, \text{or } 5\).
• "At most 2" would mean \(X = 0, 1, \text{or } 2\).
• "Exactly 2" means \(X=2\).

Using the complement rule (\(P(X \ge k) = 1 - P(X < k)\)) is often easier when \(k\) is relatively large or when the range of values for \(X \ge k\) is larger than the range for \(X < k\). In this problem with \(n=5\), calculating \(P(X=0) + P(X=1)\) (two terms) is faster than calculating \(P(X=2) + P(X=3) + P(X=4) + P(X=5)\) (four terms).