Question

15. On an average box containing 10 articles is likely to have 2 defectives. If we consider a
consignment of hundred boxes how many of them are expected to have at the most 3
defectives.
​

Answer 1

Concept:

Probability refers to possibility. A random event's occurrence is the subject of this area of mathematics. The range of the value is 0 to 1. Mathematics has incorporated probability to forecast the likelihood of various events. The degree to which something is likely to happen is basically what probability means. You will understand the potential outcomes for a random experiment using this fundamental theory of probability, which is also applied to the probability distribution. Knowing the total number of outcomes is necessary before we can calculate the likelihood that a specific event will occur.

Given:

On an average box containing 10 articles is likely to have 2 defectives.

Find:

we consider a consignment of hundred boxes how many of them are expected to have at the most 3 defectives.

Solution:

By  using Binomial distribution

P=2/10​

=0.2,n=10

since average np=2 for 10 articles.

The probability of a box having three or less defectives is P(≤3)

P(≤3)=P(0)+P(1)+P(2)+P(3)

=( ¹⁰ₓ)Pˣ(1−P)¹⁰⁻ˣ ,x=0,1,2...10

=0.879

Thus, in a consignment of 100 boxes, 0.879×100=87.90.879×100=87.9 which is approximately 88 boxes are expected to have three or less defective articles.

Therefore, the no. of boxes expected to have defective is 88

#SPJ1

Answer 2

Concept:

Probability is defined as the chances of something happening. It can be defined as the ratio of the number of favorable outcomes and the total number of outcomes.

Given:

An average box contains 10 articles, which 2 defectives are likely to have.

Find:

In the hundred boxes, expected to have at the most 3 defectives.

Solution:

Using Binomial distribution,

P=2/10=0.2,

nP=10P= 10 (0.2) = 2

nP = 2 for 10 artices.

The probability of a box having at the most 3 defectives.

P(≤3) = P(0)+P(1)+P(2)+P(3)

P(≤3) = ∑₀³(¹⁰ₓ)Pˣ (1-P)¹⁰(1-P)⁻ˣ

P(≤3) = 0.879

As there are 100 boxes in consignments,

So, P = 0.879×100 = 87.9

Hence, there are 87.9 which is 88 approx, these are expected to have at the most 3 defectives.

#SPJ3