Should add the probabilities or multiply them?

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SUMMARY

The probability of a draw in a football match between Team A and Team B, where Team A has a winning probability of 4/11 and Team B has a winning probability of 2/11, is calculated as P(draw) = 1 - (P(A wins) + P(B wins)). This results in P(draw) = 1 - (4/11 + 2/11) = 5/11. The incorrect method of multiplying the probabilities, resulting in P(draw) = 1 - (4/11)(2/11) = 113/141, is not applicable in this context. The correct approach is to recognize that the total probability of all possible outcomes must equal 1.

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Lyn05
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Hello. I'm new to this forum and I need help on this question:

Team A is playing a football match against Team B. if the probability that A will win is 4/11 and the probability that B will win is 2/11, find the probability that the match will be a draw.


I'm not sure if I should add the probabilities or multiply them, and then subtract the result from 1.


So it's either:

A) P(match will be a draw)= 1-(4/11 +2/11)
=5/11

or:

B) P(match will be a draw)= 1-(4/11)(2/11)
=113/141

Thanks a lot for any assistance provided!

Ps: Is this considered as asking for answers directly?
 
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Here is a Hint:

Let Pa be the probability that team A wins.
Let Pb be the probability that team b wins.
Let Pd be the probability of a draw.

You sample space has only three events: { A wins, B wins, Draw}

Exactly one of the three events will happen. In other words "A wins OR B wins OR Draw".

So, what can you say about Pa + Pb + Pd ?
 


Oh! So Pa + Pb + Pd = 1, therefore Pd= 1-(4/11 +2/11)
=5/11
Thank you very much!
 

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