How Do You Calculate the Probability of Drawing Two Red Balls from an Urn?

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To calculate the probability of drawing two red balls from an urn containing 3 red and 2 white balls without replacement, the solution is given by the formula \(\frac{\binom{3}{2}}{\binom{5}{2}}\). The sample space consists of all distinct combinations of drawing two balls from the five available, which totals 10 combinations. The event of drawing two red balls includes 3 combinations, corresponding to the selection of any two of the three red balls. The discussion clarifies that the sets R_1 and R_2, representing the events of drawing a red ball first and second, are not disjoint. Overall, the probability can be derived from the ratio of these combinations.
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Hi, if in an urn there are 3 red balls and 2 white balls and we draw 2 balls from the urn without replacement.
If we assume that at each ball in the urn is equally likely to be chosen, what is the probability that both balls are red?

I know the solution is \frac{\binom{3}{2}}{\binom{5}{2}}, but i want to show the elements of the sample space, for example are they the elements: r_1, r_2, r_3, w_1, w_2?
If i split the events in two disjoint events as R_1={the first ball is red} and R_{the second ball is red} what are the elements of these sets?
 
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R_1 and R_2 are not disjoint.
 
The sample space in this question contains all distinct ways of choosing two of the five balls, with 5C2 elements. Your event (in which both balls are red) has 3C2 elements, the number of distinct ways of choosing two of the three red balls.
 
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