Question

There are three categories of students in a class of 60 students: A : Very hardworking; B: Regular but not so hardworking; C: Careless and irregular 10 students are in category A, 30 in category B and rest in category C. It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is of category C​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

$\frac{200}{231}$ is the probability that this student is of category C .

Step-by-step explanation:

Explanation:

Given, there are three categories A ,B and C of students in a class of 60 students in which ,

Students in A category are very hardworking ,

Students in B category are regular but not so hardworking

Students in C category are careless and irregular .

Where 10 students  are in A , 30 students in Category B and rest students are in C .

Let ,$E_{1}$ = Selected student is of category A

    $E_{2}$ = Selected student  is of category B

$E_{3}$= Selected student is of  category C

and let X be a student could not get good marks in the examination .

Step 1:

We have total number of student is 60 .

$P(E_{1} )$ = $\frac{Number of students in category A}{Total number of students}$ = $\frac{10}{60}$  (where in A there is 10 students)

        = $\frac{1}{6}$

$P(E_{2}) = \frac{No. of student's in B}{Total number of students}$ = $\frac{30}{60}$  (where 30 students are in B category )

       = $\frac{1}{2}$

Similarly ,

$P(E_{3}) = \frac{20}{60}$ (there is 20 students in class C)

         = $\frac{1}{3}$ .

Step2:

Probability of students of category A , unable to get good marks in the final year examination is 0.002.

⇒ $P(\frac{X}{E_{1} } ) = 0.002$

Probability  of students of category B , unable to get good marks is 0.02.

⇒ $p(\frac{X}{E_{2} } ) = 0.02$

Similarly , probability of students in category C  which unable to get good marks is 0.20

⇒$P(\frac{X}{E_{3} } ) = 0.20$

Step3:

So , the probability that this student is of category C

$P(\frac{E_{3} }{X} ) = \frac{P(E_{3})P(\frac{X}{E_{3} } ) }{P(E_{1} )P(\frac{X}{E_{1} } )+P(E_{2})P(\frac{X}{E_{2} } ) + P(E_{3} )P(\frac{X}{E_{3} }) }$      (Baye's Rule )

        =               $\frac{1}{3}$  × $\frac{2}{10}$              

                  ($\frac{1}{6}$ ×$\frac{2}{1000}$) + ($\frac{1}{2}$×$\frac{2}{100}$) +($\frac{1}{3}$×$\frac{2}{10}$)

     =   $\frac{400}{2+60+400}$  = $\frac{400}{462}$

     = $\frac{200}{231}$.

Final answer :

Hence , $\frac{200}{231}$ is the probability that this student is of category C .