The game of craps is played with two dice. A player throws both dice, winning un
ID: 3353097 • Letter: T
Question
The game of craps is played with two dice. A player throws both dice, winning unconditionally if he/she produces a natural (the sum of the numbers showing on the two dice is 7 or 11), and losing unconditionally if he/she throws craps (a 2, 3, or 12). However, if the sum of the two dice is 4, 5, 6, 8, 9, or 10 (each of these is known as a point) the player continues to throw the dice until the same outcome (point) is repeated (in which case the player wins), or a 7 occurs (in which case the player loses). For example, if a player's first toss results in a 5, the player continues to roll the dice until a 5 or 7 occurs. If a 5 occurs first, the player wins, if a 7 occurs first, the player loses. (a) What is the probability that a player will toss a natural on the first roll of the dice? (b) Consider the following events for the first toss of the dice: A: Player throws craps) B: Player throws a natural C: Player throws 9, 10, or 11) (i) Which pairs of events, if any, are mutually exclusive? (ii) Which pairs of events, if any, are independent? (c) What is the probability a player throws a point on the first toss? (d) If a player throws a point on the first toss, what is the probability the player win:s the game on the next toss? (e) What is the probability that a player wins the game in two or fewer tosses?Explanation / Answer
a) Total outcomes produced as Naturals = { (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) (5,6) (6,5) }
Therefore, P( Naturals) = 8/36 = 2/9 = 0.2222
b) i) (A & B); (A & C) are mutually exclusive
ii) The events are not independent of each other as occurence of one definitely guarantees non-occurence of other event.
c) When number of outcomes are 24 it counts as points
So, P (Point) = 24/36 = 2/3 = 0.66667
d) Here the first outcome is important because, the probability of the first outcome will determine the next.
Therefore, required probability = (3/36)^2 + (4/36)^2 + (5/36)^2 + (3/36)^2 + (4/36)^2 + (5/36)^2 = 25/324 = 0.07716
e) P ( of winning in two or fewer tosses)
= P (X=1) + P(X=2)
= 2/9 + 25/324
= 0.99383
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