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gandhigauravz
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

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)


Hi ,
Can u please explain me that from where did the rolling of die B comes into picture.

There are no die B or A. There are players A and B. Player A is six spaces behind player B and the question asks about the probability that player A be on the same spot as B, after rolling a pair of fair six-sided dice. So, the question basically asks about the probability of getting 6 when rolling a pair of fair six-sided dice.

Hope it's clear.
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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)

As the question says that we roll the pair of dice together and not one by one.... thus shouldn't be (1,5) and (5,1) be considered not ordered and as one outcome rather than two different outcomes.... (same for (2,4) and (4,2)
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goodyear2013
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
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

Check out the blog posts here:
https://anaprep.com/combinatorics-sum-o ... ce-part-i/
https://anaprep.com/combinatorics-sum-o ... e-part-ii/
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@blackiscorpio : there isn't much difference between rolling 2 dice one after the other and rolling them simultaneously.. the outcome of one die is independent of the outcome of other... just like tossing two coins or tossing the same coin twice makes no difference, ( for example, probability of getting two heads is 1/2 * 1/2 , either way)

When we roll 2 dice the total number of outcomes is 36 and it includes both 2,4 and 4,2 ( and of course 5,1 and 1,5) .. these are separate independent outcomes.
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Number of possible outcomes = 6*6 = 36

Number of outcomes that result in a total of 6 (as A is 6 spaces behind B) = 5 ( (1,5), (2,4), (3,3), (4,2), (5,1) )

So, the probability = 5/36 (Option B)
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Total outcomes => 6x6 = 36

Required Outcomes => {(1,5),(2,4),(3,3),(4,2),(5,1)} => 5

Final probability => 5/36.

Hence Answer Choice B.
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Why isnt there a dice A and B here? KarishmaB Bunuel
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Why isnt there a dice A and B here? KarishmaB Bunuel

There are players A and B. The dice are not named so.

If your question is whether the dice are considered distinct, they are. Assume one is red and one is yellow.
The total possible outcomes are 36 because they are distinct and 5 of them are favorable.
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