Sam10smart wrote:
Hi. How can you ignore that Rob is also playing and that only one can win the tournament out of Mike and Ben. If Mike wins Rob Loses and so does Ben. Why wont we consider both the cases when M win B loses R loses + M loses B wins R loses?
Would appreciate your input.
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In this question, we are using the formula P(Mike or Ben) = P(Mike) + P(Ben) - P(both). Since it is stated that there can be only one champion, P(both) = 0 and thus, P(Mike or Ben) is simply the sum of the individual probabilities of success.
The reason we are ignoring Rob is that participation of Rob and his probability of success has no effect whatsoever in P(Mike or Ben). First, notice that Mike, Ben, and Rob are not the only participants in this tournament since 1/4 + 1/3 + 1/6 < 1. So not only we are ignoring Rob, we are ignoring any other participants as well. Next, the event that "Mike wins" implies that both Ben, Rob and any other participant loses. Similarly for the event "Ben wins". There are no events such as "Mike wins and Rob does not loose" or "Ben wins and Mike does not loose". That's why it is not necessary to multiply P(Mike wins) with P(Ben loses) or P(Rob loses).
In the 1-[Prob of Mike Win x Ben Lose x R lose + Prob of Mike Lose x Ben Win x R lose ] approach, you are not only assuming that Mike, Ben, and Rob are the only participants (which is not true) but also calculating as if it was possible that Mike wins and some other participant does not loose.
Hi! I dont really get the part of X does not loose... would you please elaborate? thanks
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