# An urn contains 5 red and 5 black balls. A ball is drawn at random, its color is noted and is returned to the urn. Moreover, 2 additional balls of the color drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

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Hint: Probability is the ratio of number of favorable outcomes to the number of total outcomes, use this property to reach the answer.

The urn contains 5 red and 5 black balls.

Therefore total ball becomes $ = 5 + 5 = 10$

Let a red ball be drawn in the first attempt.

$\therefore {P_1} $(Drawing a red ball) $ = \dfrac{{{\text{Red balls}}}}{{{\text{Total balls}}}}$$ = \dfrac{5}{{10}} = \dfrac{1}{2}$

If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.

So, red ball becomes $ = 7$

Therefore total ball becomes $ = 7 + 5 = 12$

$\therefore {P_2} $ (Drawing a red ball) $ = \dfrac{{{\text{Red balls}}}}{{{\text{Total balls}}}}$$ = \dfrac{7}{{12}}$

Again, the urn contains 5 red and 5 black balls.

Therefore total ball becomes $ = 5 + 5 = 10$

Now let a black ball be drawn in the first attempt.

$\therefore {P_3}$ (Drawing a black ball) $ = \dfrac{{{\text{Black balls}}}}{{{\text{Total balls}}}}$$ = \dfrac{5}{{10}} = \dfrac{1}{2}$

If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.

Therefore total ball becomes $ = 5 + 7 = 12$

$\therefore {P_4}$ (Drawing a red ball) $ = \dfrac{{{\text{Red balls}}}}{{{\text{Total balls}}}}$$ = \dfrac{5}{{12}}$

Therefore, probability of drawing second ball as red is

${P_r} = \left( {{P_1} \times {P_2}} \right) + \left( {{P_3} \times {P_4}} \right)$ (conditional probability)

${P_r} = \dfrac{1}{2} \times \dfrac{7}{{12}} + \dfrac{1}{2} \times \dfrac{5}{{12}} = \dfrac{1}{2} \times \left( {\dfrac{7}{{12}} + \dfrac{5}{{12}}} \right) = \dfrac{1}{2} \times \dfrac{{12}}{{12}} = \dfrac{1}{2}$

So the required probability is $\dfrac{1}{2}$

Note: In such types of questions the key concept we have to remember is that always remember the formula of probability which is stated above, then according to given conditions calculate different probabilities and then finally calculate your required probability using the formula which is stated above, we will get the desired answer.

The urn contains 5 red and 5 black balls.

Therefore total ball becomes $ = 5 + 5 = 10$

Let a red ball be drawn in the first attempt.

$\therefore {P_1} $(Drawing a red ball) $ = \dfrac{{{\text{Red balls}}}}{{{\text{Total balls}}}}$$ = \dfrac{5}{{10}} = \dfrac{1}{2}$

If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.

So, red ball becomes $ = 7$

Therefore total ball becomes $ = 7 + 5 = 12$

$\therefore {P_2} $ (Drawing a red ball) $ = \dfrac{{{\text{Red balls}}}}{{{\text{Total balls}}}}$$ = \dfrac{7}{{12}}$

Again, the urn contains 5 red and 5 black balls.

Therefore total ball becomes $ = 5 + 5 = 10$

Now let a black ball be drawn in the first attempt.

$\therefore {P_3}$ (Drawing a black ball) $ = \dfrac{{{\text{Black balls}}}}{{{\text{Total balls}}}}$$ = \dfrac{5}{{10}} = \dfrac{1}{2}$

If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.

Therefore total ball becomes $ = 5 + 7 = 12$

$\therefore {P_4}$ (Drawing a red ball) $ = \dfrac{{{\text{Red balls}}}}{{{\text{Total balls}}}}$$ = \dfrac{5}{{12}}$

Therefore, probability of drawing second ball as red is

${P_r} = \left( {{P_1} \times {P_2}} \right) + \left( {{P_3} \times {P_4}} \right)$ (conditional probability)

${P_r} = \dfrac{1}{2} \times \dfrac{7}{{12}} + \dfrac{1}{2} \times \dfrac{5}{{12}} = \dfrac{1}{2} \times \left( {\dfrac{7}{{12}} + \dfrac{5}{{12}}} \right) = \dfrac{1}{2} \times \dfrac{{12}}{{12}} = \dfrac{1}{2}$

So the required probability is $\dfrac{1}{2}$

Note: In such types of questions the key concept we have to remember is that always remember the formula of probability which is stated above, then according to given conditions calculate different probabilities and then finally calculate your required probability using the formula which is stated above, we will get the desired answer.

Last updated date: 17th Sep 2023

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