Soru

Zorluk: ZorTwo-Way Tables and Probability

A quality control analyst inspected a batch of 180 microchips from two production lines, Line A and Line B. The microchips were classified as Defective, Acceptable, or Premium. Some of the data from the inspection are shown in the table below.

Production LineDefectiveAcceptablePremiumTotal
Line A1050xx
Line B15yy35
Total2555180

If a microchip is selected at random from those classified as Defective or Premium, what is the probability that the chip was produced by Line A?

Cevap: 0.375

Cevap

The correct answer is 3/8 (or 0.375).
To find the probability that a randomly selected microchip was produced by Line A given that it is classified as Defective or Premium, we restrict the sample space to only the Defective and Premium microchips. From the table, the total number of Defective microchips is 25, and the total number of Premium microchips is 55, giving a combined group size of 25 + 55 = 80 microchips. Next, we determine how many of these 80 microchips were produced by Line A. Line A produced 10 Defective microchips and x Premium microchips. Since the total number of Premium microchips is 55 and Line B produced 35, Line A produced x = 55 - 35 = 20 Premium microchips. Thus, the number of microchips produced by Line A that are Defective or Premium is 10 + 20 = 30. The conditional probability is the number of favorable outcomes divided by the restricted total outcomes, which is 30/80 = 3/8 (or 0.375).

Adım Adım Çözüm

1
Find the value of xx (Line A Premium chips) using the total number of Premium chips.
x=20x = 20
Since the total number of Premium chips is 55 and Line B produced 35, Line A must have produced 5535=2055 - 35 = 20 Premium chips.
2
Calculate the total number of chips produced by Line A.
Line A Total = 80
Sum the Defective, Acceptable, and Premium chips produced by Line A: 10+50+20=8010 + 50 + 20 = 80.
3
Calculate the total number of chips produced by Line B.
Line B Total = 100
Subtract the total number of Line A chips from the grand total of 180 chips: 18080=100180 - 80 = 100.
4
Find the value of yy (Line B Acceptable chips).
y=50y = 50
Subtract the Defective (15) and Premium (35) chips of Line B from its total (100): 1001535=50100 - 15 - 35 = 50.
5
Identify the total number of chips in the conditioning category 'Defective or Premium'.
Total Defective or Premium = 80
Sum the total number of Defective chips (25) and Premium chips (55): 25+55=8025 + 55 = 80.
6
Identify the number of chips produced by Line A that are either Defective or Premium.
Favorable chips = 30
Sum the Defective chips from Line A (10) and the Premium chips from Line A (x=20x = 20): 10+20=3010 + 20 = 30.
7
Calculate the probability by dividing the favorable outcomes by the total outcomes of the conditioning category.
30/80=3/8=0.37530/80 = 3/8 = 0.375
The probability of selecting a Line A chip from the Defective or Premium group is the ratio of favorable chips to total chips in that group.

Anahtar Kavram

Conditional Probability from Two-Way Tables
Tahmini Süre:2m 30s
Bu soruyu puanla