Question

A lot consists of $12$ good pencils, $6$ with minor defects and$2$ with major defects.

A pencil is chosen at random.

The probability that this pencil is not defective is

• A. $\frac{3}{5}$
• B. $\frac{3}{10}$
• C. $\frac{4}{5}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

$\frac{3}{5}$

The explanation for the correct option:

Step1. For Option (A)

Here,

$P\left(A\right)=$ Probability of event $A$

$n\left(A\right)=$ Number of favorable events

$=$Number of good pencils

$=12$

$n\left(S\right)=$Total number of events

We know,

$\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

$\Rightarrow$ $P\left(A\right)=\frac{12}{20}$

$\Rightarrow$ $P\left(A\right)=\frac{3}{5}$[ Reducing fraction to simplest form by multiplying numerator and denominator with the highest common factor $4$]

Hence, Option (A) is the correct option.

The explanation for the incorrect options:

Step2. For Option (B)

Here, $P\left(A\right)=\frac{3}{10}$$\times \frac{2}{2}$[ Multiplying numerator and denominator with $2$]

$\Rightarrow$ $P\left(A\right)=\frac{6}{20}$

Since, $\mathbf{}\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

Here $n\left(A\right)=6,n\left(S\right)=20$

But, $n\left(A\right)=12$[Given]

Hence, Option (B) is incorrect.

Step3. For Option (C)

Here, $P\left(A\right)=\frac{4}{5}$$\times \frac{4}{4}$ [Multiplying numerator and denominator with $4$]

$\Rightarrow$ $P\left(A\right)=\frac{16}{20}$

Since, $\mathbf{}\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

Here $n\left(A\right)=16,n\left(S\right)=20$

But, $n\left(A\right)=12$[Given]

Hence, option (C) is incorrect.

Step4.For Option (D):

Here, $P\left(A\right)=\frac{1}{2}\times \frac{10}{10}$[Multiplying numerator and denominator with $10$]

$\Rightarrow$ $P\left(A\right)=\frac{10}{20}$

Since, $\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

Here $n\left(A\right)=10,n\left(S\right)=20$

But, $n\left(A\right)=12$[Given]

Hence, option (D) is incorrect.