Question

Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are summarized as follows:

shock resistance
scratch resistance high low
high 70 9
low 16 5

Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. If a disk is selected at random, determine the following probabilities. Input your answers in the fractional form (do not simplify).

(a) P(A)
(b) P(B)
(c) P(A')
(d) P(A U B)
(e) P(A' U B)

Answer 1

(a) P(A) = 86/100

(b) P(B) = 79/100

(c) P(A') = 14/100

(d) P(A U B) = 95/100

(e) P(A' U B) = 85/100

An image is attached for better understanding of the question and the table.

Explanation:

Given that;

A denote the event that a disk has high shock resistance, and

B denote the event that a disk has high scratch resistance.

(a) P(A) = N(A)/N(T)

N(A) = Total number of disks with high shock resistance.

N(T) = Total number of disks

N(A) = 70 + 16 = 86 (the first column of the table)

N(T) = 70+9+16+5 = 100

P(A) = 86/100

b) P(B) = N(B)/N(T)

N(B) = Total number of disks with high scratch resistance

N(B) = 70 + 9 = 79 (first row on the table)

N(T) = 100

P(B) = 79/100

c) P(A') = N(A')/N(T)

N(A') = Total number of disks without high shock resistance (i.e those with low shock resistance)

N(A') = 9 + 5 = 14 (second column on the table)

P(A') = 14/100

d) P(A U B) = N(AUB)/N(T)

N(AUB) = Total number of disks with high shock resistance or high scratch resistance.

N(AUB) = 70+16+9 = 95 ( add all value on the first column and row)

P(A U B) = 95/100

(e) P(A' U B) = N(A'UB)/N(T)

N(A'UB) = Total number of disks without high shock resistance or with high scratch resistance.

N(A'UB) = 70+9+5 = 84 ( second column and first row)

P(A' U B) = 84/100