Trouble with probability! HELP ME PLEASE!!!

[twobeatsoff]

There is a .97 chance that no accident will occur during any particular day at a racetrack of local importance. the probability of one accident a day is .02, and the probability of two accidents is .01.

a. What is the expected number of accidents in a day?
b. What is the expected number of accidents in ten days?

I just don't understand how to set these questions up and any help would be much appreciated.

 

[pka]

There is a .97 chance that no accident will occur during any particular day at a racetrack of local importance. the probability of one accident a day is .02, and the probability of two accidents is .01.
a. What is the expected number of accidents in a day?
b. What is the expected number of accidents in ten days?

$\displaystyle E(X)=0\cdot\mathcal{P}(X=0)+1\cdot\mathcal{P}(X=1)+2\cdot\mathcal{P}(X=2).$

 

[twobeatsoff]

$\displaystyle E(X)=0\cdot\mathcal{P}(X=0)+1\cdot\mathcal{P}(X=1)+2\cdot\mathcal{P}(X=2).$

Thank you, and I am sorry if this is asking too much, but I'm not exactly sure of what the variables your letters are supposed to represent. If you could give me a mini lesson, I would much appreciate it.

 

[pka]

Thank you, and I am sorry if this is asking too much, but I'm not exactly sure of what the variables your letters are supposed to represent. If you could give me a mini lesson, I would much appreciate it.

Expected value is the sum of the outcomes times the probability of that outcome. You have three outcomes: $\displaystyle 0,~1,~\&~2$.
Look again at my reply to see how that works.

 

[twobeatsoff]

Expected value is the sum of the outcomes times the probability of that outcome. You have three outcomes: $\displaystyle 0,~1,~\&~2$.
Look again at my reply to see how that works.

Ah! I understand perfectly now! From here, can you give me any clue as to how to set up the equation to give me the variance for part (a)?

 

[pka]

Ah! I understand perfectly now! From here, can you give me any clue as to how to set up the equation to give me the variance for part (a)?

$\displaystyle V(X)=E(X^2)-[E(X)]^2$

For $\displaystyle E(X^2)$ multiply the square of the outcome by its probability. So in this case you have to make only one change in $\displaystyle E(X)$ to get $\displaystyle E(X^2)$.