If Vahin randomly selected a permutation of the letters A, A, C, C, L, L, O, R, T, U, what is the probability he would spell “calculator”?

Answer 1

This problem can be viewed as a selection without replacement, in which we have a bag of letters (A, A, C, C, L, L, O, R, T, U) and pull them out one at a time to form the given word - in this case, we want "CALCULATOR".

As such, we need to pick out the correct letter each time, and multiply those probabilities together. Starting off, we need the probability that the first letter is C. Given that there are 10 total letters, 2 of which are C, our probability for this is 2/10, or 1/5.

Now that we've selected C as our first letter, our remaining pool shrinks to (A, A, C, L, L, O, R, T, U) - or 9 letters. The probability of the second letter being A (given that the first letter was C) is 2/9, and we can continue down this line of logic and repeatedly remove letters from the pool once they have been selected. Doing so gives us:

1. C 2/10, or 1/5
2. A 2/9
3. L 2/8, or 1/4
4. C 1/7
5. U 1/6
6. L 1/5
7. A 1/4
8. T 1/3
9. O 1/2
10. R 1/1

At the very end, we just need to multiply all of these probabilities together in order to get the answer to our question. So, we have:

1/5 * 2/9 * 1/4 * 1/7 * 1/6 * 1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/453,600.