Answer
425.1k+ views
Hint:First define each event separately using the conditional probability, and then use the Bayes formula which is given below:
Bayes formula:
\[P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\dfrac{A}{{{E_2}}}} \right)}}\]
Complete step-by-step answer:
It is given in the problem that $5\% $ of the boys and $10\% $ of the girls have an IQ of more than $150$ in a class and there are $60\% $ of boys in the class.
In this question, we need to find the probability of choosing a student at random whose IQ is more than 150 and he is a boy also.
Assume that ${E_1}$ is the event that the chosen student is a boy, as it is given that $60\% $ of the students are boys, then the probability of happening event ${E_1}$ is given as:
$P({E_1}) = \dfrac{{60}}{{100}}$
Now, assume that ${E_2}$ is the event that the chosen student is a girl, as it is given that $60\% $ of the students are boys and we also know that these two are mutually exclusive events, then the probability of happening event ${E_2}$ is given as:
$P\left( {{E_2}} \right) = 1 - P\left( {{E_1}} \right)$
$P\left( {{E_2}} \right) = 1 - \dfrac{{60}}{{100}}$
$P\left( {{E_2}} \right) = \dfrac{{100 - 60}}{{100}}$
$P\left( {{E_2}} \right) = \dfrac{{40}}{{100}}$
Let us define another event of choosing a student whose IQ is more than 150 as event A.
It is given that $5\% $ of boys have the IQ greater than $150$, then
∴ $P\left( {\dfrac{A}{{{E_1}}}} \right) = $P (a boy has an IQ of more than 150) $ = \dfrac{5}{{100}}$(given)
Similarly, it is also given that $10\% $ of girls have IQ greater than $150$, then
$P\left( {\dfrac{A}{{{E_2}}}} \right) = $P (a girl has an IQ of more than 150)$ = \dfrac{{10}}{{100}}$(given)
But we need to find the probability of the event when the chosen student with IQ more than 150 is a girl which is $P\left( {\dfrac{A}{{{E_1}}}} \right)$
Applying Bayes theorem we get,
\[P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\dfrac{A}{{{E_2}}}} \right)}}\]
Substitute the values of the probabilities into the formula:
$P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{\dfrac{{60}}{{100}} \times \dfrac{5}{{100}}}}{{\dfrac{{60}}{{100}} \times \dfrac{5}{{100}} + \dfrac{{40}}{{100}} \times \dfrac{{10}}{{100}}}}$
$P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{60 \times 5}}{{60 \times 5 + 40 \times 10}}$
$P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{300}}{{700}} = \dfrac{3}{7}$
Hence, the probability that the chosen student is the boy is \[\dfrac{3}{7}\].
Note:It can be seen that on choosing a single student in the class, there is only possibility that the chosen student is either a boy or a girl. Thus, these events are mutually exclusive and both the events do not occur together.
Bayes formula:
\[P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\dfrac{A}{{{E_2}}}} \right)}}\]
Complete step-by-step answer:
It is given in the problem that $5\% $ of the boys and $10\% $ of the girls have an IQ of more than $150$ in a class and there are $60\% $ of boys in the class.
In this question, we need to find the probability of choosing a student at random whose IQ is more than 150 and he is a boy also.
Assume that ${E_1}$ is the event that the chosen student is a boy, as it is given that $60\% $ of the students are boys, then the probability of happening event ${E_1}$ is given as:
$P({E_1}) = \dfrac{{60}}{{100}}$
Now, assume that ${E_2}$ is the event that the chosen student is a girl, as it is given that $60\% $ of the students are boys and we also know that these two are mutually exclusive events, then the probability of happening event ${E_2}$ is given as:
$P\left( {{E_2}} \right) = 1 - P\left( {{E_1}} \right)$
$P\left( {{E_2}} \right) = 1 - \dfrac{{60}}{{100}}$
$P\left( {{E_2}} \right) = \dfrac{{100 - 60}}{{100}}$
$P\left( {{E_2}} \right) = \dfrac{{40}}{{100}}$
Let us define another event of choosing a student whose IQ is more than 150 as event A.
It is given that $5\% $ of boys have the IQ greater than $150$, then
∴ $P\left( {\dfrac{A}{{{E_1}}}} \right) = $P (a boy has an IQ of more than 150) $ = \dfrac{5}{{100}}$(given)
Similarly, it is also given that $10\% $ of girls have IQ greater than $150$, then
$P\left( {\dfrac{A}{{{E_2}}}} \right) = $P (a girl has an IQ of more than 150)$ = \dfrac{{10}}{{100}}$(given)
But we need to find the probability of the event when the chosen student with IQ more than 150 is a girl which is $P\left( {\dfrac{A}{{{E_1}}}} \right)$
Applying Bayes theorem we get,
\[P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right)P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\dfrac{A}{{{E_2}}}} \right)}}\]
Substitute the values of the probabilities into the formula:
$P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{\dfrac{{60}}{{100}} \times \dfrac{5}{{100}}}}{{\dfrac{{60}}{{100}} \times \dfrac{5}{{100}} + \dfrac{{40}}{{100}} \times \dfrac{{10}}{{100}}}}$
$P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{60 \times 5}}{{60 \times 5 + 40 \times 10}}$
$P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{300}}{{700}} = \dfrac{3}{7}$
Hence, the probability that the chosen student is the boy is \[\dfrac{3}{7}\].
Note:It can be seen that on choosing a single student in the class, there is only possibility that the chosen student is either a boy or a girl. Thus, these events are mutually exclusive and both the events do not occur together.
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