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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen

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प्रश्न

There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen at random and a marble is picked from the chosen bag. What is the probability that the chosen marble is red?

बेरीज
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उत्तर

Let B1, B2, B3 be the events that bag 1, bag 2, bag 3 are selected.

Clearly P(B1) = P(B2) = P(B3) =`1/3`.

Let R1, R2, R3 be the events that red marble is drawn from bag 1, bag 2, bag 3 respectively.

If R is the event that red ball is drawn, then

R = (B1 ∩ R1) ∪ (B2 ∩ R2) ∪ (B3 ∩ R3)

The events in the brackets are mutually exclusive

∴ P(R) = P(B1 ∩ R1) + P(B2 ∩ R2) + P(B3 ∩ R3)

= `"P"("B"_1)*"P"("R"_1/"B"_1)+"P"("B"_2)*"P"("R"_2/"B"_2) + "P"("B"_3)* "P"("R"_3/"B"_3)`  ...(1)

`"P"("R"_1/"B"_1)` = Probability that red marble is drawn given that bag 1 is chosen

= `75/100       ...[("Bag 1 has 100 marbles"),("of which 75 are red")]`

Similarly `"P"("R"_2/"B"_2) = 60/100`

`"P"("R"_3/"B"_3) = 45/100`

∴ from (1), P(R) = `1/3* 75/100 + 1/3* 60/100 + 1/3 * 45/100`

=`180/300`

= `3/5`

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पाठ 9: Probability - Exercise 9.4 [पृष्ठ २०९]

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संबंधित प्रश्‍न

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Figure

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(Activity):

Mr. X goes to office by Auto, Car, and train. The probabilities him travelling by these modes are `2/7, 3/7, 2/7` respectively. The chances of him being late to the office are `1/2, 1/4, 1/4` respectively by Auto, Car, and train. On one particular day, he was late to the office. Find the probability that he travelled by car.

Solution: Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively. Let L be event that he is late.

Given that P(A) = `square`, P(C) = `square`

P(T) = `square`

P(L/A) = `1/2`, P(L/C) = `square` P(L/T) = `1/4`

P(L) = P(A ∩ L) + P(C ∩ L) + P(T ∩ L)

`="P"("A")*"P"("L"//"A") + "P"("C")*"P"("L"//"C") + "P"("T")*"P"("L"//"T")`

`= square * square + square * square + square * square`

`= square + square + square`

`= square`

`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`

= `("P"("C") * "P"("L"//"C"))/("P"("L"))`

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`= square`


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CASE-BASED/DATA-BASED
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Calculate the probability that the gems drawn are from Box II.


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