SOLUTION: An urn has 21 balls that are identical except that 5 are white, 9 are red, and 7 are blue. What is the probability that all are white if 3 are selected randomly without replacement
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Question 1204025: An urn has 21 balls that are identical except that 5 are white, 9 are red, and 7 are blue. What is the probability that all are white if 3 are selected randomly without replacement?
P = = = = 0.007519... ANSWER
The denominator is the number of all different sub-sets of 3 elements of the universal set of 21 balls.
The numerator is the number of all different sub-sets of 3 white balls of the set of 5 white balls.
The response from tutor @ikleyn shows a solution using a more sophisticated (and more powerful) method. A student just beginning to study probability should understand a more elementary way of obtaining the answer for this relatively simple problem.
The more elementary method of solving the problem is to select the three balls one at at time and consider the probability that each one selected is white.
For the first ball, there are 21 balls to choose from of which 5 are white; the probability of getting a white ball is 5/21.
For the second ball, there are 20 balls to choose from of which 4 are white; the probability of getting a white ball is 4/20.
For the third ball, there are 19 balls to choose from of which 3 are white; the probability of getting a white ball is 3/19.
The probability of all three balls being white is then the product of those individual probabilities:
