Question

 In a box of 10 electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. What is the probability that both the bulbs are without defects?
A. \[\dfrac{9}{{25}}\]
B. \[\dfrac{{16}}{{25}}\]
C. \[\dfrac{4}{5}\]
D. \[\dfrac{8}{{25}}\]

Answer 1

Hint:
First we find the probability of getting a good bulb and then we will calculate the probability of getting a defective bulb. After that, we will use the binomial distribution to calculate the required probability.

Formula Used:
\[{\rm{Probability = }}\dfrac{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{favorable}}\,{\rm{outcomes}}}}{{{\rm{The}}\,{\rm{number}}\,{\rm{of}}\,{\rm{total}}\,{\rm{outcomes}}}}\]
Binomial distribution:
\[P\left( {n,r} \right){ = ^n}{C_r}{\left( p \right)^r}{q^{n - r}}\]
\[n\]: Total number of objects selected
\[r\]: Number of chances of success
\[p\]: Probability of success
\[q\]: Probability of failure



Complete step-by-step answer:
Given that, there are 10 bulbs in a box, two are defective bulbs.
Number of good bulbs = \[10 - 2 = 8\]
Now apply the probability formula to find the probability of selecting a good bulb:
Number of favorable outcomes = 8
Total number of outcomes = 10
The probability of selecting a good bulb is \[p = \dfrac{8}{{10}}\].
\[ \Rightarrow p = \dfrac{4}{5}\]
Now apply the probability formula to find the probability of selecting a defective bulb:
Number of favorable outcomes = 2
Total number of outcomes = 10
The probability of selecting a defective bulb is \[q = \dfrac{2}{{10}}\].
\[ \Rightarrow q = \dfrac{1}{5}\]
Applying the binomial distribution to calculate the probability that both the bulbs are without defects.
The number of bulbs that pick from the box is 2.
\[P\left( {2,2} \right){ = ^2}{C_2}{\left( {\dfrac{4}{5}} \right)^2}{\left( {\dfrac{1}{5}} \right)^{2 - 2}}\]
 \[ \Rightarrow P\left( {2,2} \right) = 1{\left( {\dfrac{4}{5}} \right)^2}{\left( {\dfrac{1}{5}} \right)^0}\] Since \[^n{C_n} = 1\]
\[ \Rightarrow P\left( {2,2} \right) = \dfrac{{16}}{{25}}\]
Hence option B is the correct option.



Note:
Students are often confused in binomial distribution formulas. They do not understand which is the correct formula out of \[P\left( {n,r} \right){ = ^n}{C_r}{\left( p \right)^r}{q^{n - r}}\] or \[P\left( {n,r} \right){ = ^n}{C_r}{\left( p \right)^{n - r}}{q^r}\]. The correct formula is \[P\left( {n,r} \right){ = ^n}{C_r}{\left( p \right)^r}{q^{n - r}}\].