Example 6 from Algebra and Trigonometry, 13.7 Probability

Computing Probability Using Counting Theory

A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears.

ⓐFind the probability that only bears are chosen.
ⓑFind the probability that 2 bears and 3 dogs are chosen.
ⓒFind the probability that at least 2 dogs are chosen.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. ⓐWe need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are $C (6, 5)$ ways to choose 5 bears. There are 14 toys, so there are $C (14, 5)$ ways to choose any 5 toys.
  $\frac{C (6, 5)}{C (14, 5)} = \frac{6}{2, 002} = \frac{3}{1, 001}$
2. ⓑWe need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are $C (6, 2)$ ways to choose 2 bears. There are 5 dogs, so there are $C (5, 3)$ ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are $C (6, 2) \cdot C (5, 3)$ ways to choose 2 bears and 3 dogs. We can use this result to find the probability.
  $\frac{C (6, 2) C (5, 3)}{C (14, 5)} = \frac{15 \cdot 10}{2, 002} = \frac{75}{1, 001}$
3. ⓒIt is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen.
  When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are $C (9, 5)$ ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are $C (14, 5)$ ways to choose the 5 toys from all of the toys.
  
  $\frac{C (9, 5)}{C (14, 5)} = \frac{63}{1, 001}$
  
  If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are $C (5, 1) \cdot C (9, 4)$ ways to choose 1 dog and 1 other toy.
  $\frac{C (5, 1) C (9, 4)}{C (14, 5)} = \frac{5 \cdot 126}{2, 002} = \frac{315}{1, 001}$
  Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen.
  
  $\frac{63}{1, 001} + \frac{315}{1, 001} = \frac{378}{1, 001}$
  We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen.
  
  $1 - \frac{378}{1, 001} = \frac{623}{1, 001}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, and 2 green gumballs.

ⓐFind the probability that all 3 gumballs selected are purple.
ⓑFind the probability that no yellow gumballs are selected.
ⓒFind the probability that at least 1 yellow gumball is selected.

How did it go?

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, and 2 green gumballs.

ⓐFind the probability that all 3 gumballs selected are purple.
ⓑFind the probability that no yellow gumballs are selected.

Solution

Your turn

Solution

Your turn