Probability of stealing a base Question

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Discussion Overview

The discussion revolves around calculating the probability of at least one of three baseball players successfully stealing a base during their respective games. It involves exploring different methods for determining this probability, considering the independent nature of the events and the varying success rates of each player.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant proposes a formula for calculating the probability of at least one player stealing a base, but initially arrives at an incorrect result exceeding 100%.
  • Another participant corrects the formula, suggesting a more accurate expression that accounts for the overlaps in probabilities, resulting in a probability of 0.784.
  • A third participant presents an alternative method using the complement rule, emphasizing the independence of events and arriving at the same probability of 0.784.
  • One participant expresses appreciation for the alternative method, noting its scalability to more events.

Areas of Agreement / Disagreement

Participants generally agree on the final probability of 0.784 for at least one player stealing a base, but there are different approaches discussed to arrive at this result. The initial misunderstanding regarding the formula indicates some confusion that was later clarified.

Contextual Notes

The discussion highlights the importance of correctly applying probability formulas, particularly when dealing with independent events and overlapping probabilities. There is an implicit assumption that the players' chances of stealing bases are independent of each other.

Who May Find This Useful

This discussion may be useful for individuals interested in probability theory, particularly in the context of sports statistics and independent events. It may also benefit those looking for different methods to approach probability problems.

aj1767
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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?
 
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The formula for three items should be written as:
P(A\cupB\cupC)=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
 
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
 
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.

Pwantar said:
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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