Question

A student has to write the words ABILITY, PROBABILITY, FACILITY, MOBILITY. He wrote one word and erased all the letters in it except two consecutive letters. If 'LI' is left after erasing then the probability that the boy wrote the word PROBABILITY is: \

• A. $\frac{21}{116}$
• B. $\frac{72}{116}$
• C. $\frac{3}{5}$
• D. $\frac{4}{9}$

Hint:

Use conditional probability to find probabilities involving specific cases after an event has already occurred. Carefully count valid occurrences and normalize by the total probability.

Answer 1

The Correct Option is A

Step 1: Identifying Occurrences of 'LI' in Each Word

The given words are:

ABILITY, PROBABILITY, FACILITY, MOBILITY

We count the number of times the pair "LI" appears in each word:

ABILITY contains 'LI' once.

PROBABILITY contains 'LI' twice.

FACILITY contains 'LI' once.

MOBILITY contains 'LI' once.

Thus, the total occurrences of 'LI' in all words:

1 + 2 + 1 + 1 = 5

Step 2: Probability of Selecting Each Word

Since one word is chosen randomly, the probability of choosing any particular word is:

1/4

Step 3: Probability of 'LI' Appearing in the Chosen Word

ABILITY: 1/6 (as it has 6 consecutive letter pairs).

PROBABILITY: 2/10 = 1/5 (as it has 10 consecutive letter pairs).

FACILITY: 1/8 (as it has 8 consecutive letter pairs).

MOBILITY: 1/7 (as it has 7 consecutive letter pairs).

Step 4: Probability of Choosing 'LI' Across All Words

The total probability of 'LI' being selected is:

(1/4 × 1/6) + (1/4 × 2/10) + (1/4 × 1/8) + (1/4 × 1/7)

= 1/24 + 2/40 + 1/32 + 1/28

Converting to a common denominator of 840:

1/24 = 35/840, 2/40 = 42/840, 1/32 = 26.25/840, 1/28 = 30/840

Summing:

(35 + 42 + 26.25 + 30) / 840 = 133.25 / 840

Step 5: Computing Conditional Probability

The required probability is:

P(PROBABILITY | LI) = P(LI in PROBABILITY) / P(LI in any word)

= (1/4 × 2/10) / (133.25/840) = 21/116

Step 6: Conclusion

Thus, the final answer is:

21/116