SOLUTION: Two friends A and B play a checkers match which will end when one player has won three games (then that player is the winner) or when four games have been played (if each has won t

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Question 1183615: Two friends A and B play a checkers match which will end when one player has won three games (then that player is the winner) or when four games have been played (if each has won two games, the match is a tie). Player A is slightly better; the probability that she will win the first game is .6. When a player wins a game, her probability of winning increases by .1 (if A wins the first game, the probabilities for the second are .6 for A and .3 for B; if B wins the first game the probabilities for the second are .5 for A and .5 for B; if A wins the first two, her probability of winning the third goes up to .8, etc.)
(a) Draw the tree and calculate the probabilities for for each of the outcomes.
(b) What is the probability that A wins the match in three games?
(c) What is the probability that A wins the match at the fourth game? (there are several outcomes in this event)
(d) What is the probability that B wins the match? (e) What is the probability that the match ends in a tie?
(f) If B wins the first game, what is the probability she will win the match?
A box contains 5 red balls, 3 white balls and 2 green balls. Two balls will be taken at random, without replacement, and the colors noted (in order).
(a) If the first ball is red, what is the (conditional) probability that the second ball is also red? (You don’t need a formula for this)
(b) Draw the tree diagram for this experiment - show the probabilities on the branches (as in the model).
(c) What is the probability of getting two red balls?
(d) What is the probability of getting exactly one red ball?
Found 2 solutions by robertb, ikleyn:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
(b) P(AAA) = 0.6*0.7*0.8 = 42/125 = highlight%280.336%29

(c) P(AABA or ABAA or BAAA) = 0.6*0.7*0.2*0.7 + 0.6*0.3*0.6*0.7 + 0.4*0.5*0.6*0.7 = 273/1250 = highlight%280.2184%29

(d) P(BBB or ABBB or BABB or BBAB) = 0.4*0.5*0.6 + 0.6*0.3*0.4*0.5 + 0.4*0.5*0.4*0.5 + 0.4*0.5*0.4*0.5 = 59/250 = highlight%280.236%29

(e) P(ABAB or ABBA or BABA or BBAA or BAAB or AABB) = 0.6*0.3*0.6*0.3 + 0.6*0.3*0.4*0.5 + 0.4*0.5*0.4*0.5 + 0.4*0.5*0.4*0.5
+ 0.4*0.5*0.6*0.3 + 0.6*0.7*0.2*0.3 = 131/625 = highlight%280.2096%29

(f) P(BABB or BBAB or BBB) = 0.4*0.5*0.4*0.5 + 0.4*0.5*0.4*0.5 + 0.4*0.5*0.6 = 1/5 = highlight%280.20%29.

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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