Question

A Box Contains 10 Good Articles and 6 with Defects. One article is chosen at random. What is the Probability that It is either good or has a defect
A. \[\dfrac{{24}}{{64}}\]
B. \[\dfrac{{40}}{{64}}\]
C. \[\dfrac{{49}}{{64}}\]
D. \[\dfrac{{64}}{{64}}\]

Answer 1

Hint: Here the given question is based on the concept of probability. We have to find the probability of choosing an article of either good or defect. For this, first we need to find the total number of articles by adding both good and defect then by using the definition of probability and on further simplification by using the addition theorem of probability i.e., \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] to get the required solution.

Complete step by step answer:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.

The probability formula is defined as the probability of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
\[\text{Probability of event to happen}\, P\left( E \right) = \dfrac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\]

Given, a Box Contains 10 Good Articles and 6 with Defects.
Let \[A\] be the number of good articles \[A = 10\].
\[B\] be the number of defect articles \[B = 6\].
The total number of articles in a box \[ = 10 + 6 = 16\] articles.
The probability of choosing good articles is:
By, the definition of probability
\[ \Rightarrow \,\,P\left( \text{good articles} \right) = \dfrac{\text{Number of good articles}}{\text{Total Number of articles}}\]
\[\therefore \,\,\,P\left( A \right) = \dfrac{{10}}{{16}}\] -----(1)
The probability of choosing defect articles is:
\[ \Rightarrow \,\,P\left( \text{defect articles} \right) = \dfrac{\text{Number of defect articles}}{\text{Total Number of articles}}\]
\[\therefore \,\,\,P\left( B \right) = \dfrac{6}{{16}}\] -----(2)
The probability of choosing both good and defect articles
\[\therefore \,\,\,P\left( {A \cap B} \right) = 0\] ------(3)
Because, they randomly choose one article so there is no chance to get both the books at a time.

Now, the probability of chosen article either good or has a defect is:
Consider, the equation of addition theorem of probability:
\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\]
On substituting the values, we have
\[ \Rightarrow \,\,\,P\left( {A \cup B} \right) = \dfrac{{10}}{{16}} + \dfrac{6}{{16}} - 0\]
On simplification, we get
\[ \Rightarrow \,\,\,P\left( {A \cup B} \right) = \dfrac{{10 + 6}}{{16}}\]
\[ \Rightarrow \,\,\,P\left( {A \cup B} \right) = \dfrac{{16}}{{16}}\]
\[\therefore \,\,\,P\left( {A \cup B} \right) = 1\]
Hence, the probability of the chosen article either good or has a defect is \[1\].From the given options, \[\dfrac{{64}}{{64}}\] is equivalent to \[1\].

Therefore, option D is the correct answer.

Note:The probability is a number of possible values. Students must know the definition of the probability and their basic theorem, that is addition and multiplication theorem. \[P\left( {A \cup B} \right)\] represents the probability of occurrence of either A or B and \[P\left( {A \cap B} \right)\] represents the probability of both the event A and B will occur.