In a factory a machine produces metal bolts. In a tray of these bolts, 94% are within the allowable diameter tolerance value. The remainder exceed the tolerance. Six bolts are drawn at random from the tray. Determine the probabilities that two of the six bolts exceed the diameter.

To determine the probability that two out of six randomly drawn bolts exceed the allowable diameter tolerance value, we can use the binomial probability formula. The binomial probability formula is given by: P(X = k) = (n choose k) * p^k * (1 - p)^(n - k) Where: P(X = k) is the probability of getting exactly k successes (in this case, two bolts exceeding the tolerance). n is the total number of trials (in this case, the number of bolts drawn, which is 6). k is the number of successful trials (in this case, 2 bolts exceeding the tolerance). p is the probability of success in a single trial (the probability that a bolt exceeds the tolerance). In this scenario: n = 6 (6 bolts are drawn). k = 2 (we want to find the probability that 2 bolts exceed the tolerance). p = 1 - 0.94 (the probability that a bolt exceeds the tolerance, since 94% are within tolerance, so 6% exceed tolerance). Now, let's calculate the probability: P(X = 2) = (6 choose 2) * (0.06)^2 * (0.94)^(6 - 2) First, calculate the combinations: (6 choose 2) = 6! / (2!(6 - 2)!) = 15 Now, plug the values into the formula: P(X = 2) = 15 * (0.06)^2 * (0.94)^(6 - 2) P(X = 2) = 15 * (0.0036) * (0.94)^4 Now, calculate the final result: P(X = 2) ≈ 15 * 0.0036 * 0.797664 = 0.04310528 So, the probability that exactly two out of the six bolts drawn exceed the diameter tolerance is approximately 0.0431, or about 4.31%.