To solve the problem, we need to determine the probability that the word written by the student was "PROBABILITY," given that the letters "LI" were left after erasing the rest.

Definitions

Event B: "LI" is the sequence of letters left.

Event A: The student wrote "PROBABILITY."

We need to find $ P(A|B) $, which represents the probability of event A given event B.

Calculations

Probability of Event $ A \cap B $:

Event $ A \cap B $ happens when the student writes "PROBABILITY" and "LI" is the part remaining after erasing. In "PROBABILITY," "LI" can appear in exactly one position. Hence, the probability of choosing "PROBABILITY" and having "LI" is:

$ P(A \cap B) = \frac{1}{4} \times \frac{1}{10} $

$\frac{1}{4}$ represents the probability of choosing "PROBABILITY" from the four words.

$\frac{1}{10}$ represents the probability of choosing "LI" from the 10 possible pairs of consecutive letters in "PROBABILITY."

Similar Probability Calculations for Other Words:

For "ABILITY": "LI" can be one of 6 pairs.

$ P(\text{ABILITY}) = \frac{1}{4} \times \frac{1}{6} $

For "FACILITY": "LI" can be one of 7 pairs.

$ P(\text{FACILITY}) = \frac{1}{4} \times \frac{1}{7} $

For "MOBILITY": "LI" can be one of 7 pairs.

$ P(\text{MOBILITY}) = \frac{1}{4} \times \frac{1}{7} $

Probability of Event B:

$ P(B) = P(\text{PROBABILITY} \cap B) + P(\text{ABILITY} \cap B) + P(\text{FACILITY} \cap B) + P(\text{MOBILITY} \cap B) $

Conditional Probability $ P(A|B) $:

$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4} \times \frac{1}{10}}{\frac{1}{4} \times \frac{1}{10} + \frac{1}{4} \times \frac{1}{6} + \frac{1}{4} \times \frac{1}{7} + \frac{1}{4} \times \frac{1}{7}} $

Final Calculation:

The probability is simplified to:

$ \boxed{\frac{21}{116}} $

Probability

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