
If birth to male child and birth to female child are equal probable, then what is the probability that at least one of the three children born toa couple is male?
A. \[\dfrac{4}{5}\]
B. \[\dfrac{7}{8}\]
C. \[\dfrac{8}{7}\]
D. \[\dfrac{1}{7}\
Answer
162.6k+ views
Hint: First find the probability of birth of male child and female child. Then use the binomial distribution to obtain a function f(x). Now, substitute 1,2 and 3 for x in the obtained binomial distribution function and add \[f(1),f(2),f(3)\]to obtain the required probability.
Formula Used: The binomial distribution is,
\[f(x) = {}^n{C_x}{p^x}{q^{n - x}}\] ,
Where, f(x) is the binomial probability, x is the number of times for s specific outcome with n trials,
N is total number of trials; p is the probability of success in single trial and q is the probability of failure in single trial.
And \[{}^n{C_x} = \dfrac{{n!}}{{x!\left( {n - x} \right)!}}\] .
Complete step by step solution: Probability of male child (p)=\[\dfrac{1}{2}\] .
Probability of female child (q)= \[\dfrac{1}{2}\].
Here total number of trials is 3.
Therefore, the binomial distribution is,
\[f(x) = {}^3{C_x}{\left( {\dfrac{1}{2}} \right)^x}{\left( {\dfrac{1}{2}} \right)^{3 - x}}\]
Now, substitute 1 for x in the binomial distribution function,
\[f(1) = {}^3{C_1}{\left( {\dfrac{1}{2}} \right)^1}{\left( {\dfrac{1}{2}} \right)^{3 - 1}}\]
\[ = 3.\dfrac{1}{2}.\dfrac{1}{4}\]
=\[\dfrac{3}{8}\]
Substitute 2 for x in the binomial distribution function,
\[f(2) = {}^3{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{3 - 2}}\]
\[ = 3.\dfrac{1}{2}.\dfrac{1}{4}\]
=\[\dfrac{3}{8}\]
Substitute 3 for x in the binomial distribution function,
\[f(3) = {}^3{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{3 - 3}}\]
\[ = 1.\dfrac{1}{8}.1\]
=\[\dfrac{1}{8}\]
Add \[f(1),f(2),f(3)\] to obtain the required probability,
\[f(1) + f(2) + f(3) = \dfrac{3}{8} + \dfrac{3}{8} + \dfrac{1}{8}\]
\[ = \dfrac{7}{8}\]
Option ‘B’ is correct
Note: Sometime students get confused and only calculate f(3) and wrote the answer as \[\dfrac{1}{8}\] but here we need to calculate all the binomial distribution for 3 trials to get the answer \[\dfrac{7}{8}\] .
Formula Used: The binomial distribution is,
\[f(x) = {}^n{C_x}{p^x}{q^{n - x}}\] ,
Where, f(x) is the binomial probability, x is the number of times for s specific outcome with n trials,
N is total number of trials; p is the probability of success in single trial and q is the probability of failure in single trial.
And \[{}^n{C_x} = \dfrac{{n!}}{{x!\left( {n - x} \right)!}}\] .
Complete step by step solution: Probability of male child (p)=\[\dfrac{1}{2}\] .
Probability of female child (q)= \[\dfrac{1}{2}\].
Here total number of trials is 3.
Therefore, the binomial distribution is,
\[f(x) = {}^3{C_x}{\left( {\dfrac{1}{2}} \right)^x}{\left( {\dfrac{1}{2}} \right)^{3 - x}}\]
Now, substitute 1 for x in the binomial distribution function,
\[f(1) = {}^3{C_1}{\left( {\dfrac{1}{2}} \right)^1}{\left( {\dfrac{1}{2}} \right)^{3 - 1}}\]
\[ = 3.\dfrac{1}{2}.\dfrac{1}{4}\]
=\[\dfrac{3}{8}\]
Substitute 2 for x in the binomial distribution function,
\[f(2) = {}^3{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{3 - 2}}\]
\[ = 3.\dfrac{1}{2}.\dfrac{1}{4}\]
=\[\dfrac{3}{8}\]
Substitute 3 for x in the binomial distribution function,
\[f(3) = {}^3{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{3 - 3}}\]
\[ = 1.\dfrac{1}{8}.1\]
=\[\dfrac{1}{8}\]
Add \[f(1),f(2),f(3)\] to obtain the required probability,
\[f(1) + f(2) + f(3) = \dfrac{3}{8} + \dfrac{3}{8} + \dfrac{1}{8}\]
\[ = \dfrac{7}{8}\]
Option ‘B’ is correct
Note: Sometime students get confused and only calculate f(3) and wrote the answer as \[\dfrac{1}{8}\] but here we need to calculate all the binomial distribution for 3 trials to get the answer \[\dfrac{7}{8}\] .
Recently Updated Pages
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

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals 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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

NEET 2025 – Every New Update You Need to Know
