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05 Feb 2014, 10:44
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60% (01:42) correct
40%
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A certain board game is played by rolling a pair of fair six-sided dice and then moving one's piece forward the number of spaces indicated by the sum showing on the dice. A player is "frozen" if her opponent's piece comes to rest in the space already occupied by her piece. If player A is about to roll and is currently six spaces behind player B, what is the probability that player B will be frozen after player A rolls?
A. 1/12
B. 5/36
C. 1/6
D. 1/3
E. 17/36
OE
A. 1/12
B. 5/36
C. 1/6
D. 1/3
E. 17/36
OE
In this game a player is "frozen" whenever another player lands on the spot where the first player already has her piece. Then the question asks for the probability that player B will be frozen by player A, who's 6 spaces behind. What does it depend on? It depends on whether player A gets a roll of exactly 6 on the dice. So what's the probability of getting 6 on the dice?
First, find total possible outcomes for the roll. Since 6 numbers on each die, total number of possible rolls = 6 x 6 = 36.
There are 36 possible rolls. How many of them would allow player A to move exactly 6 spots?
Well, any roll that adds up to 6. So we could have 1/5, 2/4 and 3/3. But remember that since there are 2 dice, there are 2 ways to get 1/5 and 2/4 because either die could get the 1 or the 5, etc.
So there's 1/5, 5/1, 2/4, 4/2 and 3/3. That's 5 rolls that add up to exactly 6.
So 5 out of the 36 possible rolls would allow player A to "freeze" player B, which gives a probability of 5/36
First, find total possible outcomes for the roll. Since 6 numbers on each die, total number of possible rolls = 6 x 6 = 36.
There are 36 possible rolls. How many of them would allow player A to move exactly 6 spots?
Well, any roll that adds up to 6. So we could have 1/5, 2/4 and 3/3. But remember that since there are 2 dice, there are 2 ways to get 1/5 and 2/4 because either die could get the 1 or the 5, etc.
So there's 1/5, 5/1, 2/4, 4/2 and 3/3. That's 5 rolls that add up to exactly 6.
So 5 out of the 36 possible rolls would allow player A to "freeze" player B, which gives a probability of 5/36
05 Feb 2014, 11:01
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no. of possible outcomes = 6*6 = 36
no. of outcomes that result a total of 6 (as A is 6 spcaes behind B) = 5 ( (1,5), (2,4), (3,3), (4,2), (5,1) )
So, the probability = 5/36 (Option B)
no. of outcomes that result a total of 6 (as A is 6 spcaes behind B) = 5 ( (1,5), (2,4), (3,3), (4,2), (5,1) )
So, the probability = 5/36 (Option B)
22 Jun 2014, 18:44
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gandhigauravz
no. of possible outcomes = 6*6 = 36
no. of outcomes that result a total of 6 (as A is 6 spcaes behind B) = 5 ( (1,5), (2,4), (3,3), (4,2), (5,1) )
So, the probability = 5/36 (Option B)
no. of outcomes that result a total of 6 (as A is 6 spcaes behind B) = 5 ( (1,5), (2,4), (3,3), (4,2), (5,1) )
So, the probability = 5/36 (Option B)
Hi ,
Can u please explain me that from where did the rolling of die B comes into picture.
