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 0.01What is the probability that in a given shift there will be at most 2 defects? 0.90 0.75 0.40 0.10 1.00
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 0.01What is the probability that in a given shift there will be at most 2 defects? 0.90 0.75 0.40 0.10 1.00
Solution 1
The probability that there will be at most 2 defects in a given shift is found by adding the probabilities of having 0, 1, or 2 defects.
From the chart, we can see that:
P(X = 0) = 0.50 P(X = 1) = 0.25 P(X = 2) = 0.15
So, P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.50 + 0.25 + 0.15 = 0.90
Therefore, the probability that there will be at most 2 defects in a given shift is 0.90.
Solution 2
The probability of having at most 2 defects is the sum of the probabilities of having 0, 1, or 2 defects.
From the chart, we can see that:
P(X = 0) = 0.50 P(X = 1) = 0.25 P(X = 2) = 0.15
So, P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.50 + 0.25 + 0.15 = 0.90
Therefore, the probability that there will be at most 2 defects in a given shift is 0.90.
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