In urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it isreplaced in the urn along with 2 other balls of the same color. (a) Compute the probability that the first 2 balls selected are black and the next 2 are white. (b) Compute the probability that of the first 4 balls selected, exactly 2 are black.

a) 4.56%

b) 27.34%

Step-by-step explanation:

Let W be the white balls

Let B be the black balls

n (W) = 5

n (B) = 7

Total number of balls = 5+7 = 12

Each time a ball is drawn, it is replaced with 2 of the same color.

Let W1 be the event that the first ball drawn is white.

Let W2 be the event that the second ball drawn is white.

Let W3 be the event that the third ball drawn is white.

Let W4 be the event that the fourth ball drawn is white.

a) P (W'1W'2W3W4)

P (W'1) = 7/12

If the first ball drawn is black, the number of balls is greater by 2. So also is the number of black balls. The second black ball can be

P (W'2|W'1) = (7+2) / (12+2) = 9/14

If we already choose two black balls, the number of balls increased by 4. The number of black balls also increase by 4. There are still 5 white balls

P (W3|W'2W'1) = 5 / (12+4) = 5/16

If the event W'1, W'2 and W3 occur, there are 7 white balls in the urban and 18 balls in total

P (W4|W3W'2W'1) = (5+2) / (16+2)

=7/18

Applying multiplication principles,

P (W'1W'2W3W4) = (7*9*5*7) / (12*14*16*18)

= 0.0456

= 4.56%

b) Everytime a ball is drawn, the number of balls increases by 2.

The number of possible drawings if the ball is 12*14*16*18

The first white balls can be drawn in 5 ways and the second in 7 ways. The first black ball can be chosen in 7 ways and the second in 9 ways.

Therefore, there are 5*7*7*9 ways to choose 2 white and 2 black balls.

There are 4C2 ways i. e 6 ways of combination to draw 2 white and 2 black balls in different orders

P (2 white, 2 black balls) =

(6*5*7*7*9) / (12*14*16/18)

= 0.2734

= 27.34%