If . from Mathematics Probability Class 12 Jharkhand Board
Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Let A be the event ‘the number on the card drawn is even’ and B be the event ‘the number on the card drawn is greater than 3’.
therefore space space space space straight S space equals space open curly brackets 1 comma space 2 comma space 3 comma space 4 comma space 5 comma space 6 comma space 7 comma space 8 comma space 9 comma space 10 close curly brackets
therefore space space straight A space equals space open curly brackets 2 comma space 4 comma space 6 comma space 8 comma space 10 close curly brackets comma space space space straight B space equals space open curly brackets 4 comma space 5 comma space 6 comma space 7 comma space 8 comma space 9 comma space 10 close curly brackets
and space space space space space space straight A space intersection thin space straight B space equals open curly brackets 4 comma space 6 comma space 8 comma space 10 close curly brackets
therefore space space space space space space space straight P left parenthesis straight A right parenthesis space equals space 5 over 10 comma space space space straight P left parenthesis straight B right parenthesis space equals space 7 over 10 space and space straight P left parenthesis straight A intersection straight B right parenthesis space equals space 4 over 10
therefore space space space space space space straight P left parenthesis straight A space left enclose space straight B right parenthesis end enclose space equals space fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction space equals fraction numerator begin display style 4 over 10 end style over denominator begin display style 7 over 10 end style end fraction equals space 4 over 7.
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In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?

Let E denote the event that a student chosen randomly studies in class XII and F be the event that the randomly chosen student is a girl.
Total number of students = 1000
Number of girls = 430
Number of girls studying in class XII = 10 over 100 cross times 430 space equals space 43
Now, P(F) = 430 over 1000 space equals space 0.43  and straight P left parenthesis straight E intersection straight F right parenthesis space equals 43 over 1000 space equals 0.043
therefore space space space straight P left parenthesis straight E space left enclose straight F right parenthesis space equals space fraction numerator straight P left parenthesis straight E space intersection thin space straight F right parenthesis over denominator straight P left parenthesis straight F right parenthesis end fraction equals space fraction numerator 0.043 over denominator 0.43 end fraction space equals fraction numerator begin display style 43 over 1000 end style over denominator begin display style 43 over 100 end style end fraction equals space 43 over 100 cross times 100 over 43 space equals 1 over 10 space equals space 0.1
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If straight P left parenthesis straight A right parenthesis space equals space 7 over 13 comma space straight P left parenthesis straight B right parenthesis space equals space 9 over 13 comma space space straight P left parenthesis straight A intersection straight B right parenthesis space equals 4 over 13 comma space evaluate space straight P left parenthesis straight A space left enclose straight B right parenthesis.


Here, straight P left parenthesis straight A right parenthesis space equals 7 over 13 comma space space straight P left parenthesis straight B right parenthesis space equals space 9 over 13 comma space space straight P left parenthesis straight A intersection straight B right parenthesis space equals space 4 over 13                        ...(1)
Now,     straight P left parenthesis straight A space left enclose straight B right parenthesis space equals space fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction space equals space fraction numerator begin display style 4 over 13 end style over denominator begin display style 9 over 13 end style end fraction space space space space space space space space space space space space space space open square brackets because space space of space space left parenthesis 1 right parenthesis close square brackets
                    equals space 4 over 13 cross times 13 over 9 space equals 4 over 9.

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A and B are two events such that P(A) ≠ 0. Find P (B|A), if
(i) A is a subset of B      (ii) A ∩ B = ϕ

(i) Since A is a subset of B
therefore space space space space space straight A space intersection thin space straight B space equals space straight A                                    ...(1)
Now straight P left parenthesis straight B space left enclose straight A right parenthesis end enclose space equals space fraction numerator straight P left parenthesis straight B intersection straight A right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction space equals space fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction space equals space fraction numerator straight P left parenthesis straight A right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction                     open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
therefore space space space space straight P left parenthesis straight B space left enclose straight A right parenthesis end enclose space equals space 1

(ii) straight P left parenthesis straight B space left enclose straight A right parenthesis end enclose space equals space fraction numerator straight P thin space left parenthesis straight B intersection straight A right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction space equals space fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction

                   equals space fraction numerator straight P left parenthesis straight ϕ right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space space straight A space intersection straight B space equals straight ϕ space left parenthesis given right parenthesis close square brackets
space equals space fraction numerator 0 over denominator straight P left parenthesis straight A right parenthesis end fraction space equals space 0
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12 cards, numbered 1 to 12, are placed in a box, mixed up thoroughly and then a card is drawn at random from the box. If it is known that the number on the drawn card is more than 3, find the probability that it is an even number.

Let A be the event ‘the number on the card drawn is even’ and B be the event ‘the number on the card is greater than 3’.
therefore space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight S space equals space open curly brackets 1 comma space 2 comma space 3 comma space 4 comma space 5 comma space 6 comma space 7 comma space 8 comma space 9 comma space 10 comma space 11 comma space 12 close curly brackets
therefore space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight A space equals space open curly brackets 2 comma space 4 comma space 6 comma space 8 comma space 10 comma space 12 close curly brackets comma space straight B space equals space open curly brackets 4 comma space 5 comma space 6 comma space 7 comma space 8 comma space 9 comma space 10 comma space 11 comma space 12 close curly brackets
and             straight A space intersection thin space straight B space equals space open curly brackets 4 comma space 6 comma space 8 comma space 10 comma space 12 close curly brackets

therefore space space space space space space space straight P left parenthesis straight A right parenthesis space equals space 6 over 12 comma space space straight P left parenthesis straight B right parenthesis space equals space 9 over 12 space space and space space straight P left parenthesis straight A intersection straight B right parenthesis space equals space 5 over 12
Required probability = straight P left parenthesis straight A space left enclose straight B right parenthesis end enclose space equals space fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction space equals fraction numerator begin display style 5 over 12 end style over denominator begin display style 9 over 12 end style end fraction space equals 5 over 9
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