MHB What was the probability of the Roosters winning their last game at home?

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The Roosters have an 80% win rate at home and a 40% win rate away, with 55% of their games played at home this season. The overall probability of winning a game is calculated to be 62%. To determine the probability that their last win was at home, the conditional probability formula is applied. The intersection of winning at home and the total winning probability confirms the calculation. Thus, the probability that the Roosters' last win was at home is established through these statistical methods.
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The Roosters know that they will win 80% of their home matches and 40% of their away matches. This season’s fixture has the Roosters playing 55% of their games at home. Given that the Roosters won their last game, what was the probability that it was played at home? To find the probability of winning = 0.8x.55+0.4x0.45 = 0.62

To find if it was played at home 'given' a win then = the intersection of the two / 0.62
 
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Yes it is correct.
In this case, the conditional probability can be equated as:
$$\frac{P(winning-at-home)}{P(winning)}$$.
 
Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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