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Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 black balls. One ball is transferred from bag I to bag II and then a ball is drawn from bag II at random. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

1 Answer

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  • According to Bayes Theorem, if $E_1, E_2, E_3.....E_n$ are a set of mutually exclusive and exhaustive events, then $P\left(\large \frac{E_i}{E}\right ) = \Large \frac{P\left(\frac{E}{E_i}\right ). P(E_i)} {\sum_{i=1}^{n} (P\left(\frac{E}{E_i}\right ).P(E_i))}$
Step 1:
Given Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball drawn is red in color.
Let $E_1$ be the event that a red ball is transferred from Bag I to Bag II, and $E_2$ be the event that a black ball is transferred from Bag I to Bag II. Let A be the event that the ball drawn is red.
$P (E_1) = \large\frac{3}{3+4} = \frac{3}{7}$
$P(E_2) = \large\frac{4}{4+3} = \frac{4}{7}$
Step 2:
$P \large(\frac{A}{E_1}) = \frac{5}{10} = \frac{1}{2}$
$P \large(\frac{A}{E_2}) = \frac{4}{10} = \frac{2}{5}$
$P\left(\large \frac{E_2}{E}\right ) = \Large \frac{P\left(\frac{E}{E_2}\right ). P(E_2)} {\sum_{i=1}^{4} (P\left(\frac{E}{E_i}\right ).P(E_i))}$.
$\Rightarrow \large\;\frac{\frac{4}{7}\;\times\;\frac{2}{5}}{\large\frac{3}{7}\;\times\;\frac{1}{2}\;+\;\frac{4}{7}\;\times\;\frac{2}{5}}=\large\;\frac{16}{31}$
answered Nov 11, 2013 by sreemathi.v
 

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