The probability distribution for the number of defects during an 8-hour shift on the assembly line at Wanda’s Wooden Widgets is as shown in the chart below.x 0 1 2 3 4 5P(X = x) 0.50 0.25 0.15 0.06 0.03 0.01After 2 hours of a particular shift, 2 defects have already been detected. What is the probability that there will be less than a total of 4 defects detected during the entire shift? 1.00 0.96 0.84 0.25
Question
The probability distribution for the number of defects during an 8-hour shift on the assembly line at Wanda’s Wooden Widgets is as shown in the chart below.x 0 1 2 3 4 5P(X = x) 0.50 0.25 0.15 0.06 0.03 0.01After 2 hours of a particular shift, 2 defects have already been detected. What is the probability that there will be less than a total of 4 defects detected during the entire shift? 1.00 0.96 0.84 0.25
Solution
The probability distribution given in the question is for an 8-hour shift. However, we are asked to find the probability for the remaining 6 hours of the shift, given that 2 defects have already been detected in the first 2 hours.
First, we need to find the probability of having less than 4 defects in total during the entire shift. This means we need to find the probability of having 0 or 1 more defects in the remaining 6 hours.
The probability of having 0 more defects is P(X = 0) = 0.50. The probability of having 1 more defect is P(X = 1) = 0.25.
We add these probabilities together to get the total probability of having less than 4 defects during the entire shift:
P(X < 4) = P(X = 0) + P(X = 1) = 0.50 + 0.25 = 0.75.
Therefore, the probability that there will be less than a total of 4 defects detected during the entire shift is 0.75. This is not one of the options given in the question, so there may be a mistake in the question or the options.
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