Trouble with probability! HELP ME PLEASE!!!

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.

twobeatsoff said:

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?

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

pka said:

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

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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.

twobeatsoff said:

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.

Click to expand...

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.

pka said:

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.

Click to expand...

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)?

twobeatsoff said:

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)?

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

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 \(\displaystyle E(X^2)\).