An urn contains 10 balls of which 6 are white, 3 are red and 1 is green. Two balls are drawn at random from the urn. Find the probability distribution of number of white balls drawn. Hence state the expected number of white balls one can get in such a draw.

To find the probability distribution of the number of white balls drawn, we need to consider all possible outcomes when drawing two balls from the urn.

Let's denote W as the number of white balls drawn.

Possible outcomes when drawing two balls:
1. If both balls are white: WW
2. If one ball is white and one is red: WR or RW
3. If one ball is white and one is green: WG or GW
4. If both balls are red: RR
5. If one ball is red and one is green: RG or GR
6. If both balls are green: GG

Now, let's calculate the probabilities for each outcome:

1. Probability of drawing two white balls (WW):
The probability of drawing the first white ball is 6/10 (since there are 6 white balls out of 10 in the urn).
After drawing one white ball, there are now 5 white balls left out of 9 balls in the urn, so the probability of drawing a second white ball is 5/9.
Therefore, the probability of WW is (6/10) * (5/9) = 30/90 = 1/3.

2. Probability of drawing one white ball and one red ball (WR or RW):
The probability of drawing a white ball first is 6/10.
Since we already drew a white ball, there are now 3 red balls left out of 9 in the urn, so the probability of drawing a red ball second is 3/9.
Therefore, the probability of WR or RW is (6/10) * (3/9) = 18/90 = 1/5.

3. Probability of drawing one white ball and one green ball (WG or GW):
The probabilities are the same as case 2, so the probability of WG or GW is also 1/5.

4. Probability of drawing two red balls (RR):
The probability of drawing the first red ball is 3/10.
After drawing one red ball, there are now 2 red balls left out of 9, so the probability of drawing a second red ball is 2/9.
Therefore, the probability of RR is (3/10) * (2/9) = 6/90 = 1/15.

5. Probability of drawing one red ball and one green ball (RG or GR):
The probability of drawing a red ball first is 3/10.
After drawing a red ball, there is 1 green ball left out of 9, so the probability of drawing a green ball second is 1/9.
Therefore, the probability of RG or GR is (3/10) * (1/9) = 3/90 = 1/30.

6. Probability of drawing two green balls (GG):
The probability of drawing the first green ball is 1/10.
After drawing one green ball, there are no green balls left, so the probability of drawing a second green ball is 0/9 = 0.
Therefore, the probability of GG is (1/10) * 0 = 0.

Now, we can write down the probability distribution of the number of white balls drawn:

P(W = 0) = P(GG) = 0
P(W = 1) = P(WR or RW or WG or GW) = 1/5 + 1/5 = 2/5
P(W = 2) = P(WW) = 1/3

To find the expected number of white balls drawn, we multiply each outcome by its corresponding probability and sum them up:

E[W] = 0 * P(W = 0) + 1 * P(W = 1) + 2 * P(W = 2)
= 0 * 0 + 1 * (2/5) + 2 * (1/3)
= 2/5 + 2/3
= 6/15 + 10/15
= 16/15

Therefore, the expected number of white balls one can get in such a draw is 16/15.

To find the probability distribution of the number of white balls drawn, we can consider all possible outcomes.

Possible outcomes:
1. Two white balls are drawn.
2. One white ball and one non-white ball are drawn.
3. Two non-white balls are drawn.

Let's calculate the probabilities for each outcome:

1. Two white balls are drawn:
The probability of drawing a white ball on the first draw is 6/10.
After drawing one white ball, there are 5 white balls left out of 9 balls total.
Therefore, the probability of drawing another white ball on the second draw is 5/9.

The probability of this outcome is (6/10) * (5/9) = 30/90 = 1/3.

2. One white ball and one non-white ball are drawn:
The probability of drawing a white ball on the first draw is still 6/10.
After drawing one white ball, there are 5 non-white balls left out of 9 balls total.
Therefore, the probability of drawing a non-white ball on the second draw is 5/9.

However, there are two orders in which we can draw a white ball and a non-white ball:
- White ball on the first draw, non-white ball on the second draw.
- Non-white ball on the first draw, white ball on the second draw.

So, we need to consider both possibilities.

The probability of this outcome is (6/10) * (5/9) + (4/10) * (6/9) = 60/90 = 2/3.

3. Two non-white balls are drawn:
The probability of drawing a non-white ball on the first draw is 4/10.
After drawing one non-white ball, there are only 3 non-white balls left out of 9 balls total.
Therefore, the probability of drawing another non-white ball on the second draw is 3/9.

The probability of this outcome is (4/10) * (3/9) = 12/90 = 2/15.

Now, let's summarize the probability distribution:

Number of white balls drawn: Probability:
0 0/90 = 0
1 2/3
2 1/3

To find the expected number of white balls, we can use the formula:

Expected value = (Number of white balls * Probability) + (Number of non-white balls * Probability) + ...

Expected number of white balls =
(0 * 0) + (1 * 2/3) + (2 * 1/3) = 0 + 2/3 + 2/3 = 4/3.

Therefore, the expected number of white balls one can get in such a draw is 4/3.