Case study problem probability 4 chapter 13 class 12

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Case study  Chapter 13 (Probability )

After observing attendence register of class XII, class teacher Shri Mishra comes on conclusion that 30% students
have 100% attendence and 70% students are irregular to attend class. When he observed previous year result, he found that 70% of all students who have 100% attendence attain distinction marks while 10% irregular students attain distinction marks. At the end of the year, one student is choosen at random from the class.(Case study problem probability 4 chapter 13 class 12)

Case study problem probability 4 chapter 13 class 12
After observing attendence register of class XII, class teacher Shri Mishra comes on conclusion that 30% students

(i) The conditional probability that selected student has distinction marks given that the student has 100% attendence is
(a) 1/10                  (b) 3/10
(c) 7/10                 (d) 9/10

(ii) The conditional probability that selected student has distinction marks, given that tha student is irregular is

(a) 1/10                 (b) 3/10
(c) 7/10                 (d) 9/10

(iii) Total probability of the selected student having distinction marks from class is

(a) 4/25                  (b) 3/25
(c) 7/25                  (d) 1/25

(iv) If in random selection, selected student has distinction marks, the probability that the student has 100% attendence is

(a) 1/4                     (b) 3/4
(c) 4/7                     (d) 3/7

(v) If in random selection, selected student has distinction marks, then the probability that the student is irregular, is
(a) 3/4                    (b) 4/7
(c) 1/4                     (d) 3/7

Solution: Let X, Y and A be the event such that

X = Student has 100% attendence
Y = Student is irregular
A = Student attains distingtion marks

P(X) = 30/100, P(Y) = 70/100

P(A/X) = 70/100, P(A/Y) = 10/100

(i) Answer (c)
P( student has distinction marks given that the student has 100% attendence )

P(A/X) = 70/100 = 7/10

(ii) Answer (a)

P( student has distinction marks, given that tha student is irregular )

P(A/Y) = 10/100 = 1/10

(iii) Answer (c)

Total probability of the selected student having distinction marks

P(A) = P(X)\times P(A/X) + P(Y)\times P(A/Y)

=\frac{30}{100}\times \frac{70}{100}+ \frac{70}{100}\times \frac{10}{100}

= \frac{21}{100}+\frac{7}{100}

= \frac{28}{100}=\frac{7}{25}

(iv) Answer (b)

P( selected student has distinction marks, the probability that the student has 100% attendence)

P(X/A)= \frac{P(X)\times P(\frac{A}{X})}{P(X)\times P(\frac{A}{X})+P(Y)\times P(\frac{A}{Y})}

= \frac{\frac{30}{100}\times \frac{70}{100}}{\frac{30}{100}\times \frac{70}{100}+ \frac{70}{100}\times \frac{10}{100}}

= \frac{\frac{21}{100}}{\frac{21}{100}+\frac{7}{100}}

= \frac{21}{100}\times \frac{100}{28} = \frac{21}{28}

= \frac{3}{4}

(v) Answer (c)

P(selected student has distinction marks, then the probability that the student is irregular)

P(Y/A)= \frac{P(Y)\times P(\frac{A}{Y})}{P(X)\times P(\frac{A}{X})+P(Y)\times P(\frac{A}{Y})}

= \frac{\frac{70}{100}\times \frac{10}{100}}{\frac{30}{100}\times \frac{70}{100}+ \frac{70}{100}\times \frac{10}{100}}

= \frac{\frac{7}{100}}{\frac{21}{100}+\frac{7}{100}}

= \frac{7}{28}= \frac{1}{4}

 

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