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 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: \
Show Hint
- \( \frac{21}{116} \)
- \( \frac{72}{116} \)
- \( \frac{3}{5} \)
\( \frac{4}{9} \)
The Correct Option is A
Solution and Explanation
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
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