A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one ball is drawn from each bag, find the probability that both are green.

13/70

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1/4

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6/35

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8/35

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Solution

The correct option is C 6/35
$B a g A h a s 4 g r e e n b a l l s a n d 6 r e d b a l l s \Rightarrow p r o b a b i l i t y o f c h o o s i n g g r e e n b a l l f r o m A i s p ({g r e e n}_{A}) = \frac{4}{10} B a g A h a s 3 g r e e n b a l l s a n d 4 r e d b a l l s \Rightarrow p r o b a b i l i t y o f c h o o s i n g g r e e n b a l l f r o m B i s p ({g r e e n}_{B}) = \frac{3}{7} O n c h o o s i n g o n e b a l l f r o m e a c h b a g p r o b a b i l i t y t h a t b o t h a r e g r e e n = p ({g r e e n}_{A}) * p ({g r e e n}_{B}) = \frac{4}{10} * \frac{3}{7} = \frac{6}{35}$

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A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one ball is drawn from each bag, find the probability that both are green.

The correct option is C 6/35BagAhas4greenballsand6redballs⇒probabilityofchoosinggreenballfromAisp(greenA)=410BagAhas3greenballsand4redballs⇒probabilityofchoosinggreenballfromBisp(greenB)=37Onchoosingoneballfromeachbagprobabilitythatbotharegreen=p(greenA)∗p(greenB)=410∗37=635