SOLUTION: Hi there, I need some help with this problem In the dice game of craps two dice are tossed and the sum of the two numbers on the dice is noted. In a modified version of the g

Question 1074302: Hi there, I need some help with this problem

In the dice game of craps two dice are tossed and the sum of the two numbers on the dice is noted. In a modified version of the game a player tosses two four-sided dice. If the sum is less than four the player wins. If the sum is six or more then the player loses. If the sum is anything else, then the player tosses the two dice again and if the sum matches that of their previous sum they win, otherwise they lose.
i. Let A be the event that a player wins on their first toss of the dice, and let B be the event that a player loses on their first toss of the dice.
� Find P(A) and P(B)?
� Are events A and B mutually exclusive? Please justify your answer.
ii. If a player got a sum of four on their first throw what is the probability that they win?
iii. Find the probability that a player wins the game.

Thanks
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!


Here are the 16 possible throws of 2 4-sided dice:

(1,1) (1,2) (1,3) (1,4) 

(2,1) (2,2) (2,3) (2,4)  

(3,1) (3,2) (3,3) (3,4) 

(4,1) (4,2) (4,3) (4,4) 

So the probability of the sums are 

P(2) = P[(1,1)] = 1/16
P(3) = P[(1,2)or(2,1)] = 2/16 = 1/8
P(4) = P[(1,3)or(2,2)or(3,1)] = 3/16
P(5) = P[(1,4)or(2,3)or(3,2)or(4,1)] = 4/16 = 1/4 
P(6) = P[(2,4)or(3,3)or(4,2)] = 3/16
P(7) = P[(3,4)or(4,3)] = 2/16 = 1/8
P(8) = P((4,4)] = 1/16

P(A) = P(win on first throw) = 
P( < 4) = P(2)+P(3) = 1/16+2/16 = 3/16

P(B) = P(loses on first throw) =
P( ≥ 6) P(6)+P(7)+P(8) = 3/16+2/16+1/16 = 6/16 = 3/8

� Are events A and B mutually exclusive? Please justify your answer.

Yes, of course, because you can't both win and lose both on the 
first throw.  Think of the word "exclusive" as synonymous with
"excluding". Winning on the first throw EXCLUDES losing on the
first thown, Losing on the first throw EXCLUDES winning on the
first throw.  So the events A,B EXCLUDE each other, so they
mutually EXCLUDE each other.  So they are MUTUALLY EXCLUDING or
MUTUALLY EXCLUSIVE.

ii. If a player got a sum of four on their first throw what is the probability that they win?

P(win|4 on 1st throw) = P(4 on 2nd throw) = 3/16

iii. Find the probability that a player wins the game.

To win, one must either
1. throw less than 4
OR
2. throw 4 1st AND throw 4 2nd
OR
3. throw 5 1st AND throw 5 2nd

We remember that OR implies that we add, whereas
AND implies that we multiply, so

P(win) = 
P( < 4 on 1st throw) +
P(4 on 1st throw)�P(4 on 2nd throw) +
P(5 on 1st throw)�P(5 on 2nd throw) =
3/16 + (3/16)�(3/16) + (1/4)�(1/4) =
3/16 + 9/256 + 1/16 = 
48/256 + 9/256 + 16/256 = 73/256   

Edwin

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