Probability, rule of addition

[125mph]

Let’s say you have two players in a soccer game, each with a 25% probability to score. What is the probability that at least one will score?

SO general rule of addition for two events: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = .25 + .25 - (.25*.25) = .43.75%, correct?

How do you do the math when it is three players involved, lets say each has 25% probability to score?

Is it: P(A)+P(B)+P©-P(A and C)-P(A and B)-P(B and C)?

And for four players , is it: P(A)+P(B)+P©+P(D) -P(A and B)-P(A and C)-P(A and D)-P(B and C)- P(B and D) - P(C and D) ?

[breadmaker]

3 players: add P(A and B and C)

4 players: add all 4 combos of 3 players and subtract P(A and B and C and D)

You can also look at a Venn diagram to see it pictorially!

Or 1 - P( no one scores)

[125mph]

@breadmaker

3 players: add P(A and B and C)

4 players: add all 4 combos of 3 players and subtract P(A and B and C and D)

You can also look at a Venn diagram to see it pictorially!

Or 1 - P( no one scores)

SO you’re saying the 3 players formula is: P(A)+P(B)+P©-P(A and C)-P(A and B)-P(B and C) + [P(A and B and C)]?

And the 4 player formula is: P(A)+P(B)+P©+P(D) -P(A and B)-P(A and C)-P(A and D)-P(B and C)- P(B and D) - P(C and D) + [P(A and B and C) + P(A and B and D) + P(A and C and D) + P(B and C and D)] - [P(A and B and C and D)]?

So with each additional player, you have to add or subtract a new set?

[125mph]

@breadmaker

Or 1 - P( no one scores)

Hmm I like this shortcut route lot more.

P(A not scoring) = 1- P(A)

P(B not scoring) = 1- P(B)

P(C not scoring) = 1- P©

P(D not scoring) = 1- P(D)

Probabilty of at least one player scoring: 1 - (P(A not scoring) * P(B not scoring) * * P(C not scoring) * P(D not scoring))

Above look right?

[S2000magician]

Note that you’re assuming that the probabilities of scoring are statistically independent.

The easiest way to do these problems is to calculate the probability that nobody scores, and then subtract that from 1.

With two, the probability that nobody scores is:

(1 − 0.25)(1 − 0.25) = 0.75² = 0.5625

Therefore, the probability that at least one scores is:

1 − 0.5625 = 0.4375 = 43.75%

With three players, the probability that nobody scores is:

(1 − 0.25)(1 − 0.25)(1 − 0.25) = 0.75³ = 0.421875

and the probability that at least one scores is:

1 − 0.421875 = 0.578125 = 57.8125%

With four,

1 – 0.75⁴ = 68.359375%

And so on.

If you want to extend the addition rule to more than two events, here’s how it works:

P(A or B) = P(A) + P(B) – P(AB)

P(A or B or C) = P(A) + P(B) + P© – P(AB) – P(AC) – P(BC) + P(ABC)

P(A or B or C or D) = P(A) + P(B) + P© + P(D) – P(AB) – P(AC) – P(AD)

– P(BC) – P(BD) – P(CD) + P(ABC) + P(ABD) + P(ACD) + P(BCD) – P(ABCD)

In general, it’s:

P(At least one of n things) = Σ(probabilities taken one at a time)

– Σ(probabilities taken two at a time)

• Σ(probabilities taken three at a time)

– Σ(probabilities taken four at a time)

• ∙ ∙ ∙ – ∙ ∙ ∙ + ∙ ∙ ∙

± Σ(probabilities taken n at a time)

[125mph]

Nice work both you!

Thank you.

[breadmaker]