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Home » An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted, and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted, and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
CBSE | Class 12 | Maths | NCERT | Probability
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted, and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

There are 5 red and 5 black balls in the urn.

Let’s say the ball drawn in the first attempt is red.

\Rightarrow P (probability of drawing a red ball) =5 / 10=1 / 2

The urn now holds 7 red and 5 black balls with the addition of two balls of the same colour (red).

\Rightarrow P (probability of drawing a red ball) =7 / 12

Let’s say the ball sketched in the first try is black in colour.

\Rightarrow P (probability of drawing a black ball) =5 / 10=1 / 2

The urn now includes 5 red and 7 black balls, as well as two balls of the same colour (black).

\Rightarrow \mathrm{P} (probability of drawing a red ball) =5 / 12

As a result, the probability of drawing the second red ball is:

=\left(\frac{1}{2} \times \frac{7}{12}\right)+\left(\frac{1}{2} \times \frac{5}{12}\right)=\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)=\frac{1}{2} \times 1=\frac{1}{2}

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