If the die in Problem 7 is assumed fair, calculate the probabilities associated with the random variables in parts (a) through (d)


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Asked by 4 years ago
40 points

If the die in Problem 7 is assumed fair, calculate the probabilities associated with the random variables in parts (a) through (d).

If the
SnakeOil

1 Answer

Answered by 4 years ago
8.7k points

For these problem the total sample size is $6 \cdot 6=6^2=36$

part (a)

for P(X=1) count up the number of ways we can get a max of 1, the only ways for this to occur is for both die to roll a 1 (think about that) therefore,

$$P(X=1)=\frac{1}{36}$$

for P(X=2) count up the number of ways we can get a max of 2. We can either roll a {12,22,21}. So there are three ways of doing this and we have,

$$P(X=2)=\frac{3}{36}$$

proceed in this manner by finding P(X=3), P(X=4), P(X=5), and P(X=6) and making sure you probabilities add up to 1

part (b)

for P(X=1) think up the number of ways of getting a minimum of 1. We can roll a {11,12,13,14,15,16,61,51,41,31,21} or a total of 11 different ways to get a minimum of 1 so,

$$P(X=1) = \frac{11}{36}$$

for P(X=2) we can roll {22,23,24,25,26,62,52,42,32} or a total of 9 different ways.

$$P(X=2) = \frac{9}{36}$$

proceed in this manner making sure your probabilities add up to 1

part (c)

for P(X=2) we are interested in finding how many ways to get a sum of 2. We would have to roll a {11} hence,

$$P(X=2)=\frac{1}{36}$$

to get P(X=3) we need a sum of 3 or to roll {12,21},

$$P(X=3)=\frac{2}{36}$$

proceed in this manner making sure your probabilities add up to 1

part (d)

for P(X=-5) there is only one way to get X=-5, and that is to first roll a 1 then a 6,

$$P(X=-5)=\frac{1}{36}$$

for P(X=-4) you have to roll {26,15},

$$P(X=-4)=\frac{2}{36}$$

proceed in this manner for P(X=-3), ... , P(X=5) making sure your probabilities add up to 1

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Asked: 4 years ago

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