An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is:-

Answers (1)

Conditional Probability -

Let A and B be any two events such that B\neq \phi  or

n(B) = 0 or P(B) = 0 then P\left ( \frac{A}{B} \right ) denotes the conditional probability of occurrence of event A when B has already occured

 

-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 )

-

Let, A: Event of drawing a red ball and placing a green ball in the bag.

        B: Event of drawing a green ball and placing a red ball in the bag.

        c: Event of drawing a red ball in the second draw. 

 

P(C) = P(A)\cdot P\left(\frac{C}{A} \right ) + P(B)\cdot P\left(\frac{C}{B} \right ) \\ = \frac{5}{7}\times \frac{4}{7} + \frac{2}{7}\times \frac{6}{7} \\ = \frac{32}{49}

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