Question

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?

Answer 1

Answer:

$= \frac{1}{2}$

Step-by-step explanation:

Let E₁ and E₂ be the events that red ball is drawn in first draw and black ball is drawn in first draw respectively. Let A be the event that ball drawn in second draw is red. There are 5 red and 5 black balls in the urn.

           P(E₁) = $\frac{5}{10} = \frac{1}{2}$

         P(E₂) = $\frac{5}{10} = \frac{1}{2}$

When 2 additional balls of red colour are put in the urn there are 7 red and 5 black balls in the urn.

    ∴   P(A / E₁) = $\frac{7}{12}$

Now, When 2 additional balls of black colour are put in the urn there are 5 red and 7 black balls in the urn.

          ∴ P (A/E₂) = $\frac{5}{12}$

By theorem of total probability

       P (A) = P (E₁) P (A/E₁) + P(E₂)P(A/E₂)

       P(A) = $\frac{1}{2}$ × $\frac{7}{12}$ + $\frac{1}{2}$ × $\frac{5}{12}$

= $\frac{7}{24}$ + $\frac{5}{24}$

= $\frac{12}{24}$

= $\frac{1}{2}$

Good luck !!