A student came to an exam knowing only 20 of the 25 possible…
A student came to an exam knowing only $20$ of the $25$ possible questions. The examiner gave the student three questions. Using conditional probability, find the probability that the student knows all three questions. Determine the same probability using the classical definition of probability.
Explanation of Method
Method 1: imagine the questions are given one at a time and use the multiplication rule with conditional probabilities: each known question reduces both the number of known questions and the number of remaining questions. Method 2: count the equally likely sets of three questions, and the sets consisting only of known questions.
Step 1: Method 1 — conditional probabilities
Let $K_i$ = "the student knows the $i$-th question".
- $P(K_1)=\frac{20}{25}$;
- $P(K_2\mid K_1)=\frac{19}{24}$ (one known question used up, $24$ remain);
- $P(K_3\mid K_1K_2)=\frac{18}{23}$.
\[ P(K_1K_2K_3)=\frac{20}{25}\cdot\frac{19}{24}\cdot\frac{18}{23}=\frac{6840}{13800}=\frac{57}{115}. \]
Step 2: Method 2 — classical definition
All sets of $3$ questions out of $25$ are equally likely: $\binom{25}{3}=2300$. Favourable: sets of $3$ out of the $20$ known questions, $\binom{20}{3}=1140$.
\[ P=\frac{\binom{20}3}{\binom{25}3}=\frac{1140}{2300}=\frac{57}{115}. \]
Step 3: The answer
\[ \boxed{P=\frac{57}{115}\approx0.496.} \]
Even knowing $80\%$ of the material, the student has slightly less than an even chance of knowing all three questions.
Step 4: Why the methods agree
\[ \frac{\binom{20}{3}}{\binom{25}{3}} = \frac{20\cdot19\cdot18/3!}{25\cdot24\cdot23/3!} = \frac{20}{25}\cdot\frac{19}{24}\cdot\frac{18}{23}: \]
the $3!$ orders cancel.
Result
| Method | Computation | Result |
|---|---|---|
| conditional probability | $\frac{20}{25}\cdot\frac{19}{24}\cdot\frac{18}{23}$ | $57/115$ |
| classical | $\binom{20}3/\binom{25}3$ | $57/115\approx0.496$ |