Question

If 2% of the items of a factory are defective. The items are packed
in boxes. What is the probability that there will be (i) 2 defective
items (ii) at least3 defective items in a box of 100 items.​

Answer 1

Given:

Percentage of items defected = 2%

Total items in a box = 100

To find:

The probability that there will be

1. 2 defective  items
2. At least 3 defective items

Solution:

Number of defected items = 2% of 100 = 2/100 * 100 = 2

Probability of getting defected items =No. of Defected items/ total items

= 2/100 = 0.02

Probability of non defected items = 1 - Probability of getting defected items

= 1-0.02 = 0.98

• 2 defected items

By binomial distribution method:

We know that P(r) = ⁿCr p^r * (1-p)^(n-r)

Where n is the total number of events,

r is the number of successful events

p is the probability of successful events

(1-p) is the probability of unsuccessful events.

Probability of 2 defected item = P(2) = ¹⁰⁰C2 (0.02)² (1- 0.02)⁹⁸

= ¹⁰⁰C2 (0.02)² (0.98)⁹⁸

P(2) = 0.2734

• At least 3 defected items

At least three defected items means there must be three or more defected items:

P(at least 3) = 1 - P(0) - P(1) - P(2)

= 1 - ¹⁰⁰C₀(0.02)⁰(1- 0.02)¹⁰⁰ - ¹⁰⁰C₁(0.02)¹(1- 0.02)⁹⁹ - ¹⁰⁰C₂(0.02)²(1- 0.02)⁹⁸

1 - ¹⁰⁰C₀(0.02)⁰(0.98)¹⁰⁰ - ¹⁰⁰C₁(0.02)¹(0.98)⁹⁹ - ¹⁰⁰C₂(0.02)²(0.98)⁹⁸

= 1 - 0.1326 - 0.2707 - 0.2734

= 0.3233

The probability is 0.2734 and 0.3233 respectively.

Answer 2

Given :  2% of the items of a factory are defective.

To Find : probability that there will be (i) 2 defective

items (ii) at least3 defective items in a box of 100 items.​

Solution:

2% of the items of a factory are defective.

=> p = probability of defective = 0.02

q = probability of non-defective = 1 - 0.02 = 0.98

n = 100

P(x) = ⁿCₓpˣqⁿ⁻ˣ

(i) 2 defective  items

x = 2

=> P(2)  = ¹⁰⁰C₂(0.02)²(0.98)⁹⁸

= 0.2734

Probability  2 defective  items   = 0.2734

at least3 defective items

= 1 - P(0) - P(1) - P(2)

= 1  -   ¹⁰⁰C₀(0.02)⁰(0.98)¹⁰⁰ - ¹⁰⁰C₁(0.02)¹(0.98)⁹⁹ - ¹⁰⁰C₂(0.02)²(0.98)⁹⁸

= 1 -  0.1326 - 0.2706  -  0.2734

= 0.3594

Probability  atleast 3  defective  items   = 0.3594

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