
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
163.2k+ views
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 \[P\left( {n,r} \right){ = ^n}{C_r}{\left( p \right)^r}{q^{n - r}}\].
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 \[P\left( {n,r} \right){ = ^n}{C_r}{\left( p \right)^r}{q^{n - r}}\].
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025 Notes
