sabineodf wrote:
While I am not sure where you are calculating something wrong, I can show you my work process, I am not fantastic at this but I'm sure someone with a little bit more skill than me will have some input haha.
So, since we know that probability of something happening at least once is 1 - (probability of it not happening) we can calculate as follows
In 1 game the probability that James or Austin wins is 3/4th, so the probability that Katelyn wins in 1 game is 1/4th.
They play three games, so she has 3 chances to win: 1/4 + 1/4 + 1/4 = 3/4
Your calculation of 1 - (27/64) I believe is complicating the problem a little bit, I think you multiplied where you should have added. I did that too initially but then I remembered that her playing the game 3 times means her chances of winning increase, not decrease (as they do when you multiply).
I tried to solve this with logic as my math skills aren't great, but there are probably dozens of ways to get the right answer..
Hi
sabineodf,
In case of an
AND event, we would multiply the probabilities and not add them. The probabilities are added in case its an
OR event. To illustrate my point, let me consider the example in the question.
AND EventFor the event where Katelyn wins all 3 games, P(Katelyn winning all 3 games) = P( Katelyn winning the 1st game
AND Katelyn winning the 2nd game
AND Katelyn winning the 3rd game) \(= \frac{1}{4} * \frac{1}{4} * \frac{1}{4} = \frac{1}{64}\)
In case I extend your logic of adding the probabilities, for 5 games the probability of Katelyn winning would become
\(\frac{5}{4}\), which is not possible.OR EventExample of an OR event would be P(Katelyn winning atleast 1 game) = P(Katelyn winning only 1 game)
OR P(Katelyn winning only 2 games)
OR P(Katelyn winning all 3 games). In this case we would add the probabilities as I have done in the explanation above.
Hope its clear now!
Regards
Harsh