# A ball of unknown color was lost from an urn with 5 white and 4 red balls.

**A ball of unknown color was lost from an urn with 5 white and 4 red balls. What is the probability of taking two whites at random from the balls that are left?**

Let: event A – 2 white balls are removed from the urn (after losing the ball);

Estimated events:

H1 – the white ball was lost;

H2 – the red ball was lost;

The probability that the white ball was lost: P (H1) = 5/9; Probability that the red ball was lost: P (H2) = 4/9;

The conditional probability is to get the white balls if the white one is gone:

P (A / H1) = C (4.2) / C (8.2) the number of combinations from 4 to 2 divided by the number of combinations from 8 to 2;

C (4,2) = 4! / (2! (4 – 2)!) = 6;

C (8.2) = 8! / (2! (8 – 2)!) = 28;

P (A / H1) = 6/28 = 3/14;

The conditional probability is to get the white balls if the red one is gone:

P (A / H1) = C (5.2) / C (8.2) the number of combinations from 5 to 2 divided by the number of combinations from 8 to 2;

C (5,2) = 5! / (2! (5 – 2)!) = 10;

P (A / H2) = 10/28 = 5/14;

Total probability of event A (get 2 white balls):

P (A) = P (H1) P (A / H1) + P (H2) P (A / H2) =

= 5/9 3/14 + 4/9 5/14 = 5/42 + 10/63 = 0.2778;

Answer: The probability of taking 2 white balls from the remaining ones: P (A) = 0.2778;