Probability of stealing a base Question

1. May 7, 2008

aj1767

I have been grappling with this one for the past 24 hours.

I'll make this baseball related, consisting of three independent events assuming the following:
Player A has a 40% chance of stealing a base in "his game" tonight.
Player B has a 60% chance of stealing a base in "his game" tonight.
Player C has a 10% chance of stealing a base in "his game" tonight.

I want to figure out what the chances are that AT LEAST ONE of the three players will steal a base tonight. I originally thought the formula was P(AUBUC)=P(A)+P(B)+P(C)-P(A)*P(B)*P(C), but that gives me a final result of 1.076... a result higher than 100%. So that can't be it.

Can somebody help me out with this one?

2. May 7, 2008

ako

The formula for three items should be written as:
P(A$$\cup$$B$$\cup$$C)=P(A) + P(B) + P(C) - P(A) P(B) - P(A) P(C) - P(B) P(C) + P(A) P(B) P(C) = 0.784

3. May 7, 2008

Pwantar

Another way to do it is this:

Let S represent the number of steals.

P(S>=1) =
1 - P(S<1) =
1 - P(S=0) =
1 - P(not A and not B and not C) =
1 - P(not A)*P(not B)*P(not C) <---- Because they are independent events.

Now, P(not A) = 1-.4 = .6, P(not B) = .4, P(not C) = .9

So, P(S>=1) = 0.784

4. May 7, 2008

aj1767

Thank you both for your responses and thank you for helping me find out how to solve such problems. I like this second method since it is easier to expand to more events by simply adding more "P(not D), P(not E), etc."

Thanks again.