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If we know that team A had a $39\%$ chance of winning and team B $43\%$ chance of winning, how we can calculate the probability of the teams drawn?

My textbook mention the answer but I cannot understand the logic behind it. The answer is $18\%$. As working is not shown I guess that this is how the find $18\%$ probability of two teams withdrawn:

$$ (100\% - 39\%) - 43\% = 18\%$$

But I cannot understand the logic behind it. I appreciate if someone can explain it to me.

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4 Answers 4

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The sum of all events' probabilities is equal to 1. In this case, there are three disjoint events: team A winning, team B winning or a draw. Since we know the sum of these probabilities is 1, we can get the probability of a draw as follows:

$$ Pr(\text{Draw})=1-Pr(\text{Team A wins})-Pr(\text{Team B wins})=1-0.39-0.43=0.18 $$

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Let $A$ be the event that team A wins, let $B$ be the event team B wins, and let $D$ be the event they draw. Then since precisely one of these events must happen, $$\Pr(A)+\Pr(B)+\Pr(D)=1.$$

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The total probability of all events in a given situation must equal 1 or 100%. The probability that there is a win by EITHER team is the chance team A wins plus the chance team B wins.

This is 39% (A wins) + 43% (B wins) = 82% chance that there will be a win.

If you take the 82% away from probability of all events, 100%,

100% - 82% = 18% that the teams draw (tie)

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Either team A wins, team B wins, or they draw. The events are mutually exclusive, and for the sake of the problem, exhaust all possibilities of outcomes.

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