Probability — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and misconceptions for probability.

UT-1: Conditional Probability vs Independence Confusion

Question:

A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement.

(a) Find the probability that both balls are red.

(b) Find the probability that the second ball is red.

(c) Let $A$ be “the first ball is red” and $B$ be “the second ball is red.” A student claims that since $P(A) = \frac{5}{8}$ and $P(B) = \frac{5}{8}$The events are independent. Explain the error.

[Difficulty: hard. Tests the distinction between independent and dependent events in without-replacement scenarios.]

Solution:

(a) $P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$.

(b) By the law of total probability:

$P(B) = P(B|A)P(A) + P(B|A")P(A') = \frac{4}{7} \cdot \frac{5}{8} + \frac{5}{7} \cdot \frac{3}{8}$

$= \frac{20}{56} + \frac{15}{56} = \frac{35}{56} = \frac{5}{8}$

(c) The student’s error is applying the independence test $P(A \cap B) = P(A)P(B)$ without verifying it. In fact:

$P(A \cap B) = \frac{5}{14} \neq \frac{5}{8} \times \frac{5}{8} = \frac{25}{64}$

So $A$ and $B$ are not independent. The fact that $P(B) = P(A)$ is coincidental — it does not imply independence. The correct test for independence is $P(A \cap B) = P(A)P(B)$Not $P(B|A) = P(B)$.

UT-2: Bayes’ Theorem with Prior Identification

Question:

A factory produces items using Machine $X$ (60% of output) and Machine $Y$ (40% of output). Machine $X$ has a defect rate of 3% and Machine $Y$ has a defect rate of 5%.

(a) A randomly selected item is found to be defective. What is the probability it was produced by Machine $X$?

(b) A student reasons: “60% of items are from Machine $X$ and 3% are defective, so the probability is $0.60 \times 0.03 = 0.018$.” Identify the error.

[Difficulty: hard. Tests correct application of Bayes’ theorem versus the common prior error.]

Solution:

(a) Let $D$ be “the item is defective” and $X$ be “produced by Machine $X$.”

$P(X|D) = \frac{P(D|X)P(X)}{P(D)}$

$P(D) = P(D|X)P(X) + P(D|Y)P(Y) = 0.03 \times 0.60 + 0.05 \times 0.40 = 0.018 + 0.020 = 0.038$

$P(X|D) = \frac{0.03 \times 0.60}{0.038} = \frac{0.018}{0.038} = \frac{9}{19} \approx 0.474$

(b) The student computed $P(X \cap D)$ instead of $P(X|D)$. Bayes’ theorem asks for the posterior probability $P(X|D)$ (probability the item is from $X$ given that it is defective), but the student computed the joint probability $P(X \cap D)$ (probability the item is both from $X$ and defective, without the “given” condition). The correct computation divides by the total probability of the evidence, $P(D) = 0.038$.

Integration Tests

Tests synthesis of probability with other topics.

IT-1: Combinatorics for Probability Counting (with Number and Algebra)

Question:

A committee of 5 people is to be selected from 7 men and 5 women. The committee must contain at least 2 men and at least 2 women.

(a) In how many ways can the committee be formed?

(b) If the committee must also include exactly one person as chairperson, how many different committees are possible?

[Difficulty: hard. Combines combinatorial counting with selection under multiple constraints.]

Solution:

(a) Total ways to choose 5 from 12 people: $\binom{12}{5} = 792$.

Subtract committees with fewer than 2 men or fewer than 2 women:

• 0 men, 5 women: $\binom{7}{0}\binom{5}{5} = 1$
• 1 man, 4 women: $\binom{7}{1}\binom{5}{4} = 35$
• 2 men, 3 women: $\binom{7}{2}\binom{5}{3} = 210$
• 3 men, 2 women: $\binom{7}{3}\binom{5}{2} = 350$
• 4 men, 1 woman: $\binom{7}{4}\binom{5}{1} = 175$
• 5 men, 0 women: $\binom{7}{5}\binom{5}{0} = 21$

Total excluded: $1 + 35 + 210 + 350 + 175 + 21 = 792$.

The constraint “at least 2 men and at least 2 women” only excludes the cases with 0 or 1 of one gender:

Excluded: 0 men, 5 women (1 way) + 1 man, 4 women (35 ways) + 5 men, 0 women (21 ways) + 4 men, 1 woman (175 ways) = 232 ways.

$\text{Valid committees} = 792 - 232 = 560$

(b) For each valid committee of 5, there are 5 choices for chairperson. So: $560 \times 5 = 2800$ different chaired committees.

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