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 High LowScratch High 70 9Resistance Low 16 5Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. Are events A and B independent?

Part a

The probability that the event disk has high shock resistance is 0.86

Part b

The probability that a disk has high scratch resistance given that the disk has high shock resistance is 0.8140

Step-by-step explanation:

(a).

From the given information,

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.

                          SHOCK HIGH(A)     SHOCK LOW(A)          TOTAL

SCRATCH  

HIGH(B)               70                               9                                79

RESISTANCE       16                               5                                  21

LOW(B)

TOTAL                86                                 14                                100

Compute P(A).

Therefore, the probability value of the event A is 0.86.

Part a

The probability that the event disk has high shock resistance is 0.86

Explanation | Hint for next step

Based on the given information, the probability that the event disk has high shock resistance is 0.86. That means it is approximately equal to 86%.

Step 2 of 2

(b)

From the given information,

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.

                      SHOCK HIGH(A)     SHOCK LOW(A)            TOTAL

SCRATCH

HIGH(B)             70                               9                               79

SCRATCH

LOW(B)             16                                 5                               21

TOTAL             86                                15                               100

ComputeP(B/A) = P(A∩B) /P(A)  ;P(A) > 0

P(A∩B)  =70/100

 =0.70

From the part [a], the probability value of the event A is P(A) =0.86 .

Therefore,

P( {B/A}  = P(A∩B)  /P(A)

=  0.70 /0.86

=0.8140

Part b

The probability that a disk has high scratch resistance given that the disk has high shock resistance is 0.8140

Explanation | Common mistakes

The disk with high scratch resistance is found to be 81.40% with the condition that the disk with high shock resistance is maintained.