Probability
Probability
Allen Star Batch
Grade 12
Question:
The letters of the word PROBABILITY are written down at random in a row. Let $E_1$ denote the event that two $i$, $s$ are together and $E_2$ denote the event that two $B$'s are together, then:
$P(E_1) = P(E_2) = \frac{3}{11}$
$P(E_1 \cap E_2) = \frac{2}{55}$
$P(E_1 \cup E_2) = \frac{19}{55}$
$P(E_1 / E_2) = \frac{1}{5}$
Step-by-Step Solution
Key Concept: When finding probability of events with identical repeated elements, treat identical letters as a single unit and use the principle that P(E) = (favorable arrangements)/(total arrangements). The word PROBABILITY has 11 letters with 2 I's and 2 B's, so total arrangements = 11!/2!2! and arrangements with identical letters together = 10!/2!
Given $P(E_1) = P(E_2) = \frac{10!}{11!} \cdot \frac{2!}{2!} = \frac{2}{11}$. For both $I$'s and both $B$'s to be together, $P(E_1 \cap E_2) = \frac{9!}{11!} \cdot \frac{2!}{2!} = \frac{2}{55}$. Using inclusion-exclusion, $P(E_1 \cup E_2) = \frac{2}{11} + \frac{2}{11} - \frac{2}{55} = \frac{18}{55}$. Therefore, $P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{2/55}{2/11} = \frac{1}{5}$.
Correct Answer: 2,3,4