Consider all the arrangements of the letters from the word 'APTITUDE'.
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Consider all the arrangements of the letters from the word 'APTITUDE'. Let E1 be the event of 'all vowels together' and E2 be the event of 'all consonants together'. What is the value of P(E1 ∪ E2) [i.e. the probability of the occurrence of E1 or E2]?
Show answer & explanation
P(E1 ∪ E2) = P(E1) + P(E2) - P(E1 ∩ E2). Since all vowels cannot be together AND all consonants together simultaneously in 'APTITUDE' (which has 4 vowels and 4 consonants distributed across 8 letters), P(E1 ∩ E2) = 0. Computing P(E1) = 2/35 and P(E2) = 2/35 (derived from arrangements where vowel/consonant blocks are treated as single units), the total probability is 2/35 + 2/35 = 4/35.
Step-by-step Derivation:
Step 1: Identify letters in 'APTITUDE': A, P, T, I, T, U, D, E (8 letters total).
Step 2: Classify letters:
- Vowels: A, I, U, E (4 vowels)
- Consonants: P, T, T, D (4 consonants, with T repeated twice)
Step 3: Calculate total arrangements:
Total = 8! / 2! = 40320 / 2 = 20160 (accounting for repeated T)
Step 4: Calculate N(E1) - all vowels together:
Treat 4 vowels as 1 block + 4 consonants = 5 units
Arrangements = 5! × 4! / 2! = 120 × 24 / 2 = 1440
P(E1) = 1440 / 20160 = 2/28 = 1/14 (wait, recalculate)
Actually: 5! × 4! / 2! = 120 × 12 = 1440
P(E1) = 1440 / 20160 = 1/14
Step 5: Calculate N(E2) - all consonants together:
Treat 4 consonants as 1 block + 4 vowels = 5 units
Arrangements = 5! × 4! / 2! = 120 × 12 = 1440 (the 2! accounts for repeated T in consonant block)
P(E2) = 1440 / 20160 = 1/14
Step 6: Calculate P(E1 ∩ E2):
If all vowels are together AND all consonants are together, we have 2 blocks = 2! × 4! × 4! / 2! = 2 × 24 × 12 = 576
P(E1 ∩ E2) = 576 / 20160 = 1/35
Step 7: Apply inclusion-exclusion:
P(E1 ∪ E2) = P(E1) + P(E2) - P(E1 ∩ E2) = 1/14 + 1/14 - 1/35
= 5/70 + 5/70 - 2/70 = 8/70 = 4/35