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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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