
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
164.4k+ 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 \[\dfrac{7}{8}\] .
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