Suppose that the function f: [0, 1] →ℝ is integrable. Prove...

Suppose that the function $f: [0, 1] \rightarrow \mathbb{R}$ is integrable. Prove that \\ $\lim_{n \rightarrow \infty} \frac{1}{n} \left[ f\left(\frac{1}{n}\right) + f\left(\frac{2}{n}\right) + \dots + f\left(\frac{n-1}{n}\right) + f(1) \right] = \int_0^1 f$.

Transcript

00:01 In this problem we are told that a deck of 52 cards is dealt evenly to 4 players such that each player has 13 cards.

00:08 Now, for part a of the problem, we have to find the probability that the ace of hertz and the ace of spades are dealt to two different players.

00:16 Now, we know that 2 hertz can be dealt to 2 different players in 4 multiplied by 3 that is equal to 12 ways.

00:38 And and the rest of the 50 cards can be dealt to 4 players 12 to 2 because these two cards these 2 hertz have all sorry these two hertz have already been dealt so 50 cards will be dealt to 4 players such that 12 of them will get 12 cards each and other and the rest of the 2 will get 13 cards each in how many ways 50 factorial divided by 13 factorial square divided by 12 factorial square divided by 12 factorial square ways in these many ways all the cards can be dealt now with this we can say that the total number of ways all the 52 cards of the deck can be dealt is given by 52 factorial because there are 52 cards and 13 factorial raised to the power 4 because there are 13 cards of each of the suits that are to be dealt to four different players so with this we can say that the required probability p will be given by 12 multiplied by 50 factorial divided by 13 factorial divided by 12 factorial square whole divided by total number of ways which is 52 factorial divided by 13 factorial raised to the power 40 after simplifying this expression we get the answer to the probability approximately equal to 0 .764 and this is the answer to the first part of the problem now for the second part of the problem, part b, we have to find that, find the probability that all the four players receive one ace each given that the ace of hearts and the ace of spades are dealt to two different players...

Question

Suppose that the function $f: [0, 1] \rightarrow \mathbb{R}$ is integrable. Prove that \\ $\lim_{n \rightarrow \infty} \frac{1}{n} \left[ f\left(\frac{1}{n}\right) + f\left(\frac{2}{n}\right) + \dots + f\left(\frac{n-1}{n}\right) + f(1) \right] = \int_0^1 f$.

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