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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability Exercise 30.5 Question 23 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.5 Question 23 Maths Textbook Solution.

Answers (1)

Answer:\frac{17}{42}

Hint: You must know the rules of finding probability functions.

Given:  Urn A = (4R+3B) = 7 balls

Urn B = (5R+4B) = 9 balls

Urn C = (4R+4B) = 8 balls

Solution

Urn A contain 4 red \left ( R_{1} \right ), 3 black \left ( B_{1} \right ) balls

Urn B contain 5 red \left ( R_{2} \right ), 4 black \left ( B_{2} \right ) balls

Urn B contain 4 red \left ( R_{3} \right ), 4 black \left ( B_{3} \right ) balls

P(3 balls drawn contain 2 red and a black ball)

\begin{aligned} &=P\left[\left(R_{1} \cap R_{2} \cap B_{3}\right) \cup\left(R_{1} \cap B_{2} \cap R_{3}\right) \cup\left(B_{1} \cap R_{2} \cap R_{3}\right)\right] \\\\ &=P\left(R_{1} \cap R_{2} \cap B_{3}\right)+P\left(R_{1} \cap B_{2} \cap R_{3}\right)+P\left(B_{1} \cap R_{2} \cap R_{3}\right) \\\\ &=P\left(R_{1}\right) \times P\left(R_{2}\right) \times P\left(B_{3}\right)+P\left(R_{1}\right) \times P\left(B_{2}\right) \times P\left(R_{3}\right)+P\left(B_{1}\right) \times P\left(R_{2}\right) \times P\left(R_{3}\right) \\\\ &=\frac{4}{7} \times \frac{5}{9} \times \frac{4}{8}+\frac{4}{7} \times \frac{4}{9} \times \frac{4}{8}+\frac{3}{7} \times \frac{5}{9} \times \frac{4}{8} \\\\ &=\frac{80+64+60}{504} \\\\ &=\frac{204}{504} \\\\ &=\frac{17}{42} \end{aligned}

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