Question

two fair 6 - sided dice, one blue and one red, are tossed, and the up face on each die is recorded. define the following events: e: { the difference of the numbers is 3 or more } f: { the sum of the numbers is even } find the following probabilities: (a) p(e)= (b) p(f)= (c) p(e and f)= are events e and f independent? a. no b. yes note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. show past answers

Explanation:

Step1: Calculate total number of outcomes

When two 6 - sided dice are rolled, the total number of outcomes is $n(S)=6\times6 = 36$.

Step2: Find outcomes for event $E$

The pairs for which the difference of the numbers is 3 or more are: $(1,4),(1,5),(1,6),(2,5),(2,6),(3,6),(4,1),(5,1),(5,2),(6,1),(6,2),(6,3)$. So $n(E)=12$. Then $P(E)=\frac{n(E)}{n(S)}=\frac{12}{36}=\frac{1}{3}$.

Step3: Find outcomes for event $F$

The sum of two numbers is even if both numbers are even or both numbers are odd. The number of pairs with both odd numbers: $3\times3 = 9$ (pairs like $(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)$). The number of pairs with both even numbers: $3\times3=9$ (pairs like $(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)$). So $n(F)=9 + 9=18$. Then $P(F)=\frac{n(F)}{n(S)}=\frac{18}{36}=\frac{1}{2}$.

Step4: Find outcomes for $E\cap F$

The pairs that are in both $E$ and $F$ are: $(1,5),(5,1),(2,6),(6,2)$. So $n(E\cap F)=4$. Then $P(E\cap F)=\frac{n(E\cap F)}{n(S)}=\frac{4}{36}=\frac{1}{9}$.

Step5: Check for independence

Two events $E$ and $F$ are independent if $P(E\cap F)=P(E)\times P(F)$. $P(E)\times P(F)=\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}
eq\frac{1}{9}=P(E\cap F)$. So the events are not independent.

Answer:

(a) $\frac{1}{3}$
(b) $\frac{1}{2}$
(c) $\frac{1}{9}$
A. no