Question

It is known that a box of 600 electric bulbs contains 12 detective bulbs. One bulb is taken out at random. What is the probability that it is a non-defective bulb?

Answer 1

Hint: We solve this question by first finding the number of non-defective bulbs in the box. Then we find the number of ways in which we can draw a bulb from the box using the formula, the number of ways of selecting r objects from n objects is equal to ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$. Then we use the same formula to find the number of ways of selecting a non-defective bulb. Then we use the formula $P\left( A \right)=\dfrac{\text{Number of outcomes of A}}{\text{Total number of outcomes}}$, to find the probability that the drawn bulb is non-defective.

Complete step-by-step solution:
We are given that a box contains 600 electric bulbs. It is also given that the number of defective bulbs in those 600 bulbs is 12 bulbs.
So, the number of non-defective bulbs are equal to
$600-12=588$
So, the number of non-defective bulbs in the box is equal to 588.
We are given that a bulb is drawn at random.
Let us consider the formula, number of ways of selecting r objects from a set of n objects is
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$
Using this formula, we get the number of ways of selecting 1 bulb from 600 bulbs is
\[\begin{align}
  & \Rightarrow {}^{600}{{C}_{1}}=\dfrac{600!}{1!\times \left( 600-1 \right)!} \\
 & \Rightarrow {}^{600}{{C}_{1}}=\dfrac{600!}{1!\times 599!}=600 \\
\end{align}\]
So, number of ways of selecting one bulb from the box is 600.
Now let us find the number of ways of selecting a non-defective bulb.
Let us consider the formula, number of ways of selecting r objects from a set of n objects is
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$
Using this formula, we get the number of ways of selecting 1 non-defective bulb from 588 bulbs is
\[\begin{align}
  & \Rightarrow {}^{588}{{C}_{1}}=\dfrac{588!}{1!\times \left( 588-1 \right)!} \\
 & \Rightarrow {}^{588}{{C}_{1}}=\dfrac{588!}{1!\times 587!}=588 \\
\end{align}\]
So, number of ways of selecting 1 non-defective bulb from 588 bulbs is 588.
Now we need to find the probability that it is a non-defective bulb.
Now, let us consider the formula for probability.
The probability of any event A can be given as,
$P\left( A \right)=\dfrac{\text{Number of outcomes of A}}{\text{Total number of outcomes}}$
Using the formula, we can write the probability of drawing a non-defective bulb as,
$P\left( \text{selecting non-defective bulb} \right)=\dfrac{\text{Number of ways of selecting 1 non-defective bulb}}{\text{Total number of ways of selecting 1 bulb}}$
So, by substituting the above obtained values in the formula we get,
$\begin{align}
  & \Rightarrow P\left( \text{selecting non-defective bulb} \right)=\dfrac{\text{588}}{\text{600}} \\
&\Rightarrow P\left( \text{selecting non-defective bulb} \right)=\dfrac{\text{49}}{\text{50}}\\
\end{align}$
So, the probability of drawing a non-defective bulb is $\dfrac{\text{49}}{\text{50}}$.
Hence, the answer is $\dfrac{\text{49}}{\text{50}}$.

Note: The common mistake made while solving this problem is one goes in a hurry and solves the answer for probability of drawing a defective bulb as
$\begin{align}
  & \Rightarrow P\left( \text{selecting defective bulb} \right)=\dfrac{\text{12}}{\text{600}} \\
 & \Rightarrow P\left( \text{selecting defective bulb} \right)=\dfrac{1}{\text{50}} \\
\end{align}$
Then they get the answer as $\dfrac{1}{\text{50}}$. So, one must read the question carefully and understand what we are asked to find.