How Likely Is a Jockey to Fall in the Next Ten Rides?

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The discussion revolves around calculating the probability of a jockey falling at least once in the next ten rides, given that she falls once every ten rides on average. The correct probability of falling at least once is determined to be 0.651. Participants explore the approach of summing probabilities for different fall scenarios but realize that this can lead to incorrect results if not done properly. A more efficient method is suggested, focusing on the complementary probability of not falling at all. The conversation emphasizes understanding the total probability distribution for the outcomes of falls during the rides.
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Homework Statement


On average a certain jockey falls from her horse once in every 10 rides. What is the probability that she will fall from her horse at least once in the next ten rides?


Homework Equations


Answer is 0.651.


The Attempt at a Solution


I've no idea .. I thought of adding up the probabilities of the jockey falling once, twice, thrice, etc. but that just gives me an answer greater than 1 which would not make sense.

P(Fall at least once) = (1/10) + (1/10 + 1/10) + (1/10 + 1/10 + 1/10) + ...
= 55/10
 
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You want to add up the probability that the jockey falls once out of the ten jumps+ the probability that the jockey falls twice out of the ten +... The probability that the jockey falls once is equal to 0.1*0.9*0.9*...*0.9 (where there are nine 0.9's).

Summing this for all the possibilities (i.e. 1 to 9 falls) will give the answer. However, can you see a quick way of doing this? What must the sum of the probability of all the choices (i.e. falls no times, falls once, falls twice...) be?
 
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