Of the tools produced with a certain manufacturing process, 10% are defective. Using: a) the binomial distribution and b) the Poisson approximation to the binomial distribution, find the probability that in a sample of 10 tools chosen at random exactly 2 are defective.

a) Using the Binomial Distribution: In this case, we have a binomial distribution with the following parameters: Number of trials (n): 10 (sample size). Probability of success (p) for each trial (defective tool): 0.10 (10% are defective). Probability of failure (q) for each trial (non-defective tool): 1 - 0.10 = 0.90. We want to find the probability of getting exactly 2 defective tools, which can be calculated using the binomial probability formula: $$P(X=k)=nC_k\cdot{p^k}\cdot{q^{n−k}}$$ Where: P(X=k) is the probability of getting exactly k defective tools. n is the number of trials (10). k is the number of successes (2). p is the probability of success (0.10). q is the probability of failure (0.90). Now, plug in the values and calculate: $$P(X=2)=10C2\cdot(0.10)^2(0.90)^8$$ Using the binomial coefficient $$P(X=2)=45\cdot(0.10)^2(0.90)^8$$ Calculate the result: P(X=2)=0.1937 So, using the binomial distribution, the probability of getting exactly 2 defective tools in a sample of 10 is approximately 0.1937.