Probability of stealing a base Question

  • Thread starter aj1767
  • Start date
  • #1
2
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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?
 

Answers and Replies

  • #2
ako
1
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The formula for three items should be written as:
P(A[tex]\cup[/tex]B[tex]\cup[/tex]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
5
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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
2
0
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.

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
 

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