The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by $p$. Then $100p =$ _________.

We are given:

• Total number of computers $= 10$

• Working computers $= 4$

• Non-working computers $= 6$

• Number of computers inspected $= 4$

The total number of ways to choose 4 computers out of 10 is:

The system is deemed functional if at least 3 of the 4 inspected computers are working. The possible favorable cases are:

1. Exactly 3 working computers and 1 non-working computer:

   $$\binom{4}{3} \times \binom{6}{1} = 4 \times 6 = 24$$

2. All 4 working computers and 0 non-working computers:

   $$\binom{4}{4} \times \binom{6}{0} = 1 \times 1 = 1$$

Total favorable outcomes $= 24 + 1 = 25$.

The probability $p$ is:

Therefore:

This falls within the official range of 11.85 to 11.95.