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

[bjgawp]

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

 

[cristo]

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?

 

[Architeuthis Dux]

= 5.5

As a scientist, you are correct in thinking that adding up the probabilities of the jockey falling once, twice, thrice, etc. would result in a value greater than 1. However, this is because you are not taking into account the fact that these events are not mutually exclusive. In other words, the jockey could fall more than once in the next ten rides, and your calculation does not account for that.

To find the probability that the jockey will fall at least once in the next ten rides, we can use the complementary probability approach. This means finding the probability that the jockey will not fall at all in the next ten rides, and then subtracting that value from 1.

P(Fall at least once) = 1 - P(No falls in 10 rides)

To find the probability of no falls in 10 rides, we can use the binomial distribution formula:

P(No falls in 10 rides) = (10 choose 0) * (1/10)^0 * (9/10)^10
= 1 * 1 * (9/10)^10
= 0.3486784401

Therefore, the probability that the jockey will fall at least once in the next ten rides is:

P(Fall at least once) = 1 - 0.3486784401
= 0.6513215599

This is the same answer you provided, and it is correct. The probability that the jockey will fall at least once in the next ten rides is approximately 0.651 or 65.1%.