A recent study has shown that a new treatment cures a certain disease $\$ 1 \%$ of the time. A random sample of 11 patients undergoing this treatment is chosen. Find the probability that fewer than 10 of them are cured.
Do not round your intermediate computations, and round your answer to three decimal places.

Step 1 ：This problem is a binomial distribution problem. The binomial distribution model is appropriate for a statistical experiment if the following conditions are met: 1. The experiment consists of n repeated trials. 2. Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. 3. The probability of success, denoted by P, is the same on every trial. 4. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.

Step 2 ：In this case, we have n=11 trials (patients), each trial can result in two outcomes (cured or not cured), the probability of success (cured) is 0.01, and the trials are independent (one patient being cured does not affect whether another patient is cured).

Step 3 ：The probability mass function of a binomial distribution is given by: \[ P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)) \] where: - P(X=k) is the probability of k successes in n trials - C(n, k) is the number of combinations of n items taken k at a time - p is the probability of success on a single trial - n is the number of trials - k is the number of successes

Step 4 ：We need to find the probability that fewer than 10 patients are cured, which means we need to find P(X<10). This is equal to the sum of P(X=k) for k=0 to 9.

Step 5 ：Given n = 11 and p = 0.01, the probability that fewer than 10 out of 11 patients are cured is 1.0, or 100%. This makes sense because the probability of a patient being cured is very low (1%), so it is highly likely that fewer than 10 patients will be cured.

Step 6 ：Final Answer: The probability that fewer than 10 out of 11 patients are cured is \(\boxed{1.000}\).