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Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise 30.5 Question 20 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer:\frac{43}{90}

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

Given:

Bag \left ( A \right )\rightarrow \left ( 4R+5B \right )balls

Bag \left ( B \right )\rightarrow \left ( 3R+7B \right )balls

Solution:

Total no. of balls = 9 balls in bag A

Total no. of balls =10 balls in bag B

 P(balls of different colors) =

 P (red from bag A and black from bag B) + P (red from bag B + black from bag A)

=\left ( \frac{4}{9}\times \frac{7}{10} \right )+\left ( \frac{3}{10}\times \frac{5}{9} \right )

=\frac{28}{90}+\frac{15}{90}

=\frac{43}{90}

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infoexpert21

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