An accountant is going to carry out the audit of 24 accounts of a company. 16 of these accounts correspond to high volume clients. If the accountant randomly selects 4 accounts, what is the probability that at least one is high volume? Recommendation: First find the probability that none of the accounts are high volume.

1. Calculate the total number of ways to select 4 accounts out of 24:
\binom{24}{4} = \frac{24!}{4!(24-4)!} = \frac{24!}{4! \cdot 20!}

2. Calculate the number of ways to select 4 low volume accounts out of 8:
\binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8!}{4! \cdot 4!}

3. Compute the probability of selecting none of the high volume accounts:
P(\text{none}) = \frac{\binom{8}{4}}{\binom{24}{4}} = \frac{\frac{8!}{4!4!}}{\frac{24!}{4!20!}}

4. Compute the probability of selecting at least one high volume account:
P(\text{at least one}) = 1 - \frac{\binom{8}{4}}{\binom{24}{4}}

Simplify to get the final probability:
P(\text{at least one}) = 1 - \frac{70}{10626} = 1 - 0.00659 = 0.99341

Therefore, the probability that at least one of the selected accounts is a high volume account is:

P(\text{at least one}) \approx 0.99341