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There are two urns. There are m white & n black balls in the first urn and p white & q black balls in the second urn. One ball is taken from the first urn & placed into the second. Now, the probability of drawing a white ball from the second urn is:

Option: 1

\frac{pm + (p+1)n}{(m+n)(p+q +1)}


Option: 2

\frac{ (p+1)m+ pn }{(m+n)(p+q +1)}


Option: 3

\frac{qm + (q+1)n}{(m+n)(p+q +1)}


Option: 4

\frac{(q+1)m+ qn}{(m+n)(p+q +1)}


Answers (1)

best_answer

   

Conditional Probability -

 

P\left ( \frac{A}{B} \right )= \frac{P\left ( A\cap B \right )}{P\left ( B \right )}

and

P\left ( \frac{B}{A} \right )= \frac{P\left ( A\cap B \right )}{P\left ( A \right )}

 

- wherein

where P\left ( \frac{A}{B} \right ) probability of A when B already happened.

 

 

Independent events -

If A and B are independent events then probability of occurrence of A is not affected by occurrence or non occurrence of event B.

\therefore P\left ( \frac{A}{B} \right )= P\left ( A \right )

and    \dpi{100} \therefore P\left ( A\cap B \right )= P\left ( B \right )\cdot P\left ( \frac{A}{B} \right )

so  \therefore P\left ( A\cap B \right )= P\left ( A \right )\cdot P \left ( B \right )= P\left ( AB \right )

-

 

 

  A ball from first urn can be drawn is two mannars

            ball is white     or  ball is black

         p (w) = m / (m+n)           p (B) = n / (m+n)

      Let E \rightarrow selecting a white ball from second urn after a ball from urn first has been placed into it

            P(E) = P(w) P(E/W) + P(B) P(E/B)      

= \frac{m }{m+n} \times \frac{p+1 }{p+q+1} + \frac{n}{m+n}\frac{p }{p+q+1 } \\\\

\frac{ (p+1)m+ pn }{(m+n)(p+q +1)}

 

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Nehul

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