Hello, margaritas!

I can help with part (a) . . . I'm still working on part (b).

We have the following probabilitues:A bag contains 2 red marbles, 5 white marbles and 3 black marbles. .A game is played

with a player randomly drawing one marble at a time from the bag with replacement,

continuing until he obtains a red marble or a white marble.

He wins if the last marble is white and loses otherwise.

(a) Find the probability that the player wins.

. . P(W) = 0.5 . wins

. . P(R) = 0.2 . loses

. . P(B) = 0.3 . draws again

Let's baby-step through this . . .

He could win on the first draw: .P(W1) .= .0.5

He could win on the second draw.

. . He must draw B on the first draw: P(B1) = 0.3

. . then draw W on the second draw: P(W2) = 0.5

Hence: .P(W2).= .(0.5)(0.3)

He could win on the third draw.

. . He must draw B on the first draw: P(B1) = 0.3

. . then draw B on the second draw: P(B2) = 0.3

. . then draw W on the third draw: P(W3) = 0.5

Hence: .P(W3) .= .(0.5)(0.3)²

and so on . . .

Then the probability that he wins is:

. . P(win) .= .0.5 + (0.5)(0.3) + (0.5)(0.3)² + (0.5)(0.3)³ + . . .

This is a geometric series with first term a = 0.5 and common ratio r = 0.3

. . . . . . . . . . . . . . .0.5 . . . . 0.5

Its sum is: .S .= . --------- .= .----

. . . . - . . . . . . . . 1 - 0.3 . - . 0.7

Therefore: .P(win) .= .5/7