Basic Engineering Mathematics, Fifth Edition

Probability 309

4. Determine the probability that the total score
   is 8 when two like dice are thrown.


5. The probability of eventAhappening is

3

5

and the probability of eventBhappening is

2

3

. Calculate the probabilities of

(a) bothAandBhappening.

(b) only event Ahappening, i.e. event A

happening and eventBnot happening.

(c) only eventBhappening.

(d) eitherA,orB,orAandBhappening.

6. When testing 1000 soldered joints, 4 failed
   during a vibration test and 5 failed due to
   having a high resistance. Determine the prob-
   ability of a joint failing due to
   (a) vibration.
   (b) high resistance.
   (c) vibration or high resistance.
   (d) vibration and high resistance.

Here are some further worked problems on probability.

Problem 6. A batch of 40 components contains 5

which are defective. A component is drawn at

random from the batch and tested and then a second

component is drawn. Determine the probability that

neither of the components is defective when drawn

(a) with replacement and (b) without replacement

(a) With replacement

The probabilitythatthe component selected on the

first draw is satisfactory is 35/40 i.e. 7/8. The com-

ponentis nowreplaced and a second draw is made.

The probability that this component is also satis-

factory is 7/8. Hence, the probability that both the

first component drawn and the second component

drawn are satisfactory is

7

8

×

7

8

=

49

64

or 0.7656

(b) Without replacement

The probability that the first component drawn

is satisfactory is 7/8. There are now only 34

satisfactory components left in the batch and the

batch number is 39. Hence, the probability of

drawing a satisfactory component on the sec-

ond draw is 34/39. Thus, the probability that the

first component drawn and the second component

drawn are satisfactory i.e., neither is defective is

7

8

×

34

39

=

238

312

or 0.7628

Problem 7. A batch of 40 components contains 5

which are defective. If a component is drawn at

random from the batch and tested and then a second

component is drawn at random, calculate the

probabilityof having one defective component, both

(a) with replacement and (b) without replacement

The probabilityof having one defective component can

be achieved in two ways. Ifpis the probabilityof draw-

ing a defective component andqis the probability of

drawing a satisfactory component, then the probability

of having one defective component is given by drawing

a satisfactory component and then a defective compo-

nentorby drawing a defective component and then a

satisfactory one; i.e., byq×p+p×q.

(a) With replacement

p=

5

40

=

1

8

and q=

35

40

=

7

8

Hence, the probability of having one defective

component is

1

8

×

7

8

+

7

8

×

1

8

i.e.

7

64

+

7

64

=

7

32

or 0.2188

(b) Without replacement

p 1 =

1

8

andq 1 =

7

8

on the first of the two draws

The batch number is now 39 for the second draw,

thus,

p 2 =

5

39

and q 2 =

35

39

p 1 q 2 +q 1 p 2 =

1

8

×

35

39

+

7

8

×

5

39

=

35 + 35

312

=

70

312

or 0.2244