The word ACTIVATED consists of the letters A, C, T, I, V, A, T, E, D. There are 2 As and 2 Ts.
We need to find the probability that the selection of 5 letters does not contain more Ts than As.
Method 1: Summing number of ways
- AT___: Choose 2 As and 1 T, then choose 2 more letters from the remaining 5 letters: \(2 \times 2 \times \binom{5}{3} = 40\)
- A____: Choose 1 A, then choose 4 more letters from the remaining 7 letters: \(2 \times \binom{7}{4} = 10\)
- AATT_: Choose 2 As and 2 Ts, then choose 1 more letter from the remaining 5 letters: \(\binom{5}{1} = 5\)
- AAT__: Choose 2 As and 1 T, then choose 2 more letters from the remaining 5 letters: \(2 \times \binom{5}{2} = 20\)
- AA___: Choose 2 As, then choose 3 more letters from the remaining 7 letters: \(\binom{5}{3} = 10\)
- _____: Choose 5 letters from the remaining 7 letters: \(\binom{5}{5} = 1\)
\(Total number of ways not containing more Ts than As = 86\)
Probability = \(\frac{86}{\binom{9}{5}} = \frac{43}{63} \approx 0.683\)
Method 2: Subtracting number of ways with more Ts from total
- T____: Choose 1 T, then choose 4 more letters from the remaining 7 letters: \(2 \times \binom{7}{4} = 10\)
- TTA__: Choose 2 Ts and 1 A, then choose 2 more letters from the remaining 5 letters: \(2 \times \binom{5}{2} = 20\)
- TT___: Choose 2 Ts, then choose 3 more letters from the remaining 7 letters: \(\binom{5}{3} = 10\)
\(Total number of ways with more Ts than As = 40\)
Total number of ways to choose 5 letters from 9 = \(\binom{9}{5} = 126\)
Number of ways not containing more Ts than As = \(126 - 40 = 86\)
Probability = \(\frac{86}{126} = \frac{43}{63} \approx 0.683\)
