In a certian lottery, 4 differnt numbers between 1 and 11 inclusive are drawn. These are the winning numbers. A player selects 4 different numbers between 1 and 11. What is the probability that the player has all winning numbers?

To find the probability that the player has all winning numbers, we need to determine the total number of possible outcomes and the number of favorable outcomes.

In this case, the total number of possible outcomes can be calculated by using the concept of combinations. Since the player selects 4 different numbers between 1 and 11, the total number of possible outcomes is given by the formula:

nCr = n! / (r!(n-r)!)

Where n is the total number of numbers (11 in this case) and r is the number of numbers the player selects (4 in this case).

So, the total number of possible outcomes is:

11C4 = 11! / (4!(11-4)!) = 330

Now, to calculate the number of favorable outcomes (the player getting all winning numbers), since there are only 4 winning numbers, the player must select these 4 specific numbers from the pool of 11 numbers. Therefore, the number of favorable outcomes is:

4C4 = 1

Now, we have the total number of possible outcomes (330) and the number of favorable outcomes (1). The probability of the player having all the winning numbers can be calculated as:

Probability = Number of Favorable Outcomes / Total Number of Possible Outcomes

Probability = 1/330

Hence, the probability that the player has all the winning numbers is 1/330.

there are C(11,4) or 330 ways to draw 4 numbers from 11

One of these is the winning combination.
So the probability is ..... ?