2 probability Q's. Help please!

[mdenham2]

2 probability questions... please explain how you arrived at your answer if possible--thanks!

(1) A fair die is rolled 8 times. What is the probability of getting:

a) 1 on each of the eight rolls?

b) 6 exactly twice in the eight rolls?

c) 6 at least once in the eight rolls?


(2) In a car race, there are 6 Chevrolets, 4 Fords, and 2 Pontiacs. In how many ways can the 12 cars finish if we consider only the make of the cars?

 

[soroban]

Hello, mdenham2!

(1) A fair die is rolled 8 times.
What is the probability of getting:

a) 1 on each of the eight rolls?

\(\displaystyle \text{The probability of getting a 1 on one roll is: }\,\tfrac{1}{6}\)

\(\displaystyle P(\text{eight 1's}) \:=\:\left(\frac{1}{6}\right)^8 \:=\:\frac{1}{1,679,616}\)

b) 6 exactly twice in the eight rolls?

\(\displaystyle P(\text{two 6's}) \;=\;{8\choose2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^6 \;=\;\frac{109,375}{419,904}\)

c) 6 at least once in the eight rolls?

The opposite of "at least one 6" is "no 6's."

\(\displaystyle P(\text{no 6's}) \;=\;\left(\frac{5}{6}\right)^8 \;=\;\frac{390,625}{1,679,616}\)

\(\displaystyle \text{Therefore: }\(\text{at least one 6}) \;=\;1 - \frac{390,625}{1,679,616} \;=\;\frac{1,288,991}{1,679,616}\)

(2) In a car race, there are 6 Chevrolets, 4 Fords, and 2 Pontiacs.
In how many ways can the 12 cars finish if we consider only the make of the cars?

\(\displaystyle \frac{12!}{6!\,4!\,2!} \;=\;13,860\)

 

[mdenham2]

Thank you for your kind help. It is much very appreciated.

-Matt