Question

stat 213 assignment 1: problem 2 (1 point) two fair 6 - sided dice are tossed, and the up - face on each die is recorded. find the probability of observing each of the following events: a: { the sum of the numbers is 10 or more } b: { a 5 does not appear on either die } c: { the sum of the numbers is odd } p(a)= p(b)= p(c)= note: you can earn partial credit on this problem.

Explanation:

Step1: Find total number of outcomes

When two 6 - sided dice are tossed, the total number of outcomes is $n(S)=6\times6 = 36$ since each die has 6 possible outcomes.

Step2: Find outcomes for event A

The pairs of numbers on the two dice such that their sum is 10 or more are: $(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)$. So $n(A)=6$. Then $P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}$.

Step3: Find outcomes for event B

The numbers on a 6 - sided die are 1, 2, 3, 4, 5, 6. The pairs where 5 does not appear on either die: There are 5 choices for each die (1, 2, 3, 4, 6), so $n(B)=5\times5 = 25$. Then $P(B)=\frac{n(B)}{n(S)}=\frac{25}{36}$.

Step4: Find outcomes for event C

For the sum of two numbers to be odd, one number must be even and the other must be odd. If the first die is even (3 choices: 2, 4, 6) and the second is odd (3 choices: 1, 3, 5), there are $3\times3 = 9$ outcomes. If the first die is odd (3 choices: 1, 3, 5) and the second is even (3 choices: 2, 4, 6), there are $3\times3=9$ outcomes. So $n(C)=9 + 9=18$. Then $P(C)=\frac{n(C)}{n(S)}=\frac{18}{36}=\frac{1}{2}$.

Answer:

$P(A)=\frac{1}{6}$
$P(B)=\frac{25}{36}$
$P(C)=\frac{1}{2}$