Drawing/Selecting/Picking/Choosing Two/Three/More fruits/lamps/bulbs/objects/things from a Box/Bag/Urn

Problem 4

A lot of 100 bulbs from manufacturing process is known to contains 10 defective and 90 non defective bulbs. If a sample of 8 bulbs is selected at random. What is the probability that a) the sample has 3 defective and 5 non – defective b) the sample has at least one defective bulb?

Solution

Total number of bulbs in the lot

= 10 Defective + 90 Non-Defective
= 100

Number of bulbs selected = 8

Experiment : Selecting 8 bulbs from the lot

Total Number of Possible Choices

= Number of ways in which the 8 bulbs can be selected from the total 100 bulbs
⇒ n = 100C8
=
100 × 99 × 98 × 97 × 96 × 95 × 94
8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 10 × 33 × 7 × 97 × 19 × 94

Let

  • A be the event of the sample having 3 defectives and 5 non-defectives
  • B be the of the sample having at least one defective

For Event A

[Defectives] [Non-Defectives] Total
Available 10 90 100
To Choose 3 5 8
Choices 10C390C5100C8

Number of Favorable Choices

= The number of ways in which the 3 defective and 5 non-defective
bulbs can be selected from the total 100 bulbs
⇒ mA = (Number of ways in which 3 defective bulbs can be selected from the 10)
× (Number of ways in which 5 non-defective bulbs can be selected
      from the 90)
Fundamental Counting Theorem (of Multiplication): Where an event can be sub divided into two or more independent sub-events, the total number of ways in which the total event can be accomplished is equal to the product of the number of ways in which the sub-events can be accomplished.
= 10C3 × 90C5
=
10 × 9 × 8
3 × 2 × 1
×
100 × 99 × 98 × 97 × 96
5 × 4 × 3 × 2 × 1
= 120 × 5 × 33 × 49 × 97 × 96

Probability of the sample having 3 defectives and 5 non-defectives

⇒ Probability of occurrence of Event A

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(A) =
mA
n
=
10C3 × 90C5
100C8

For Event B [not preferred]

Fundamental Counting Theorem (of Addition): Where an event can be accomplished in a number of alternative ways, the total number of ways in which the event can be accomplished is equal to the sum of the number of ways in which the alternative events can be accomplished.

Event B can be accomplished in eight alternative ways.

  • B1 : Selecting 1 defective and 7 non-defectives
  • B2 : Selecting 2 defectives and 6 non-defectives
  • B3 : Selecting 3 defectives and 5 non-defectives
  • B4 : Selecting 4 defectives and 4 non-defectives
  • B5 : Selecting 5 defectives and 3 non-defectives
  • B6 : Selecting 6 defectives and 2 non-defectives
  • B7 : Selecting 7 defectives and 1 non-defectives
  • B8 : Selecting 8 defectives and 0 non-defectives
Since this method needs a lot of calculations to be made, the alternative route of finding the probability for this event using the complimentary event of getting all defectives is adopted.

For Event B

Probability of at least one of the bulbs being defective

= 1 − Probability of none of the bulbs being defective
P(B) = 1 − P(C)
=
1 −
45
100C8
=
100C8 − 45
100C8

Let C be the event of all the bulbs being defective

For Event C

[Defectives] [Non-Defectives] Total
Available 10 90 100
To Choose 8 0 8
Choices 10C890C0100C8

Number of Favorable Choices

= The number of ways in which the 8 defective bulbs can
be selected from the total 100 bulbs
⇒ mC = 10C8
= 10C(10 − 8)nCr = nC(n − r)
= 10C2
=
10 × 9
2 × 1
= 45

Probability of none of the bulbs being defective

⇒ Probability of occurrence of Event C

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(C) =
mC
n
=
45
100C8
Author : The Edifier