Question

Two dice are tossed (with 6 faces each) at the same time. Define the outcomes X1=’ result from dice #1’ and X2=’ result from dice #2’. Now, consider M= X1-X2 , where M is given by the difference between the outcomes. a) What is the probability that M is negative ? b) What is the probability that M is positive ? c) What is the probability that M=0 ? d) If X2=1 what is the probability that M=6 ? e) If M=1 what is the probability that X1=6 ?

Answer 1

Answer:

$(a)\frac{5}{12} (b)\frac{5}{12} (c)\frac{1}{6} (d)0 (e)\frac{1}{36}$

Step-by-step explanation:

The pairs of the Outcome from the two dice $(X_1,X_2)$ are given below:

$(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)\\(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)\\(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)\\(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)\\(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)\\(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)$

Given $M=X_1-X_2 \text{, M=Difference between the outcomes}$, the sample space of M is:

$0, -1, -2,-3,-4,-5 \\1, 0, -1, -2,-3,-4\\2, 1, 0, -1, -2,-3\\3,2, 1, 0, -1, -2\\4,3,2, 1, 0, -1\\5, 4,3,2, 1, 0$

(a)Probability that M is negative

n(Negative Terms)=15

$P( M is negative )=\dfrac{15}{36}= \dfrac{5}{12}$

(b)Probability that M is positive

n(Positive Terms)=15

$P( M is positive )=\dfrac{15}{36}= \dfrac{5}{12}$

(c)Probability that M=0

n(Zeroes)=6

$P( M=0 )=\dfrac{6}{36}= \dfrac{1}{6}$

(d) n(M=6)=0

$P(M=6|X_2=1)=\frac{0}{36}=0$

(e)If M=1, the probability that $X_1=6$

When $X_1=6$, the number of outcomes where M=1 is 1.

Therefore:

$P(X_1=6|M=1)=\frac{1}{36}$