MHB Probability of drawing a 10 on the 4th ball?

[Yankel]

Hello, I have this question, I need some help with it.

In an urn there are 10 balls marked with the numbers 1-10.
4 balls are being chosen randomly without replacement.
It is known that on the 4th ball that was taken out, there was a number which was larger than the number on any of the previous 3 balls taken.
What is the probability that on the 4th ball the number was 10 ?

I thought of conditional probability, the intersection should be in my opinion:

(9/10)*(8/9)*(7/8)*(1/7)

I am not sure it's right and anyway I don't know what to do with the probability of the 4th being larger than the other 3...

thanks !

 

[soroban]

Hello, Yankel!

This is Conditional Probability.
Are you familiar with Bayes' Theorem? .$P(A\,|\,B) \;=\;\dfrac{P(A \wedge B)}{P(B)}$
If not, it is useless to solve the problem for you.

In an urn there are 10 balls marked with the numbers 1 - 10.
4 balls are being chosen randomly without replacement.
It is known that the 4th ball had a number larger than the number on any of the previous 3 balls.
What is the probability that on the 4th ball the number was 10 ?

 

[Plato2]

In an urn there are 10 balls marked with the numbers 1-10.
4 balls are being chosen randomly without replacement.
It is known that on the 4th ball that was taken out, there was a number which was larger than the number on any of the previous 3 balls taken.
What is the probability that on the 4th ball the number was 10 ?

Let us suppose that $K$ is the value of the fourth ball that is greater than any of the first three drawn. Clearly $K=4,~5,\cdots,10$.
It should be clear that $\mathcal{P}(K=4)=\dfrac{3!\cdot 6!}{10!}$. Yes or NO?

What is $\mathcal{P}(K=n)$ for $n=5,~6,\cdots,10~?$