JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 23)
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 :
$${{21} \over {49}}$$
$${{27} \over {49}}$$
$${{26} \over {49}}$$
$${{32} \over {49}}$$
Explanation
5 Red and 2 green balls
P(one red ball) = $${5 \over 7}$$
P(one green ball) = $${2 \over 7}$$
Case I :
If drawn ball is green than a red ball is added
$$\left( {\matrix{ {6{\mathop{\rm Re}\nolimits} d} \cr {1\,Green} \cr } } \right)$$ P (red ball) = $${6 \over 7}$$
Case II :
If drawn ball is red than a green ball is added
$$\left( {\matrix{ {4{\mathop{\rm Re}\nolimits} d} \cr {3\,Green} \cr } } \right)$$ P (red ball) = $${4 \over 7}$$
P (2nd red ball) = $${5 \over 7}$$ $$ \times {4 \over 7} + {2 \over 7} \times {6 \over 7}$$ = $${{32} \over {49}}$$
P(one red ball) = $${5 \over 7}$$
P(one green ball) = $${2 \over 7}$$
Case I :
If drawn ball is green than a red ball is added
$$\left( {\matrix{ {6{\mathop{\rm Re}\nolimits} d} \cr {1\,Green} \cr } } \right)$$ P (red ball) = $${6 \over 7}$$
Case II :
If drawn ball is red than a green ball is added
$$\left( {\matrix{ {4{\mathop{\rm Re}\nolimits} d} \cr {3\,Green} \cr } } \right)$$ P (red ball) = $${4 \over 7}$$
P (2nd red ball) = $${5 \over 7}$$ $$ \times {4 \over 7} + {2 \over 7} \times {6 \over 7}$$ = $${{32} \over {49}}$$
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