13.7 Probability (Page 4/18)

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Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10.

$\frac{5}{6}$

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Computing probability using counting theory

Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems.

Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are $C (5, 2)$ ways to select 2 phones that are not defective. There are 8 phones, so there are $C (8, 2)$ ways to select 2 phones. The probability of selecting 2 phones that are not defective is:

\begin{array}{l} \frac{ways to select 2 phones that are not defective}{ways to select 2 phones} = \frac{C (5, 2)}{C (8, 2)} \\ = \frac{10}{28} \\ = \frac{5}{14} \end{array}

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.

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.
$\frac{C (6, 5)}{C (14, 5)} = \frac{6}{2, 002} = \frac{3}{1, 001}$
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.
$\frac{C (6, 2) C (5, 3)}{C (14, 5)} = \frac{15 \cdot 10}{2, 002} = \frac{75}{1, 001}$
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.

$1 - \frac{378}{1, 001} = \frac{623}{1, 001}$

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Questions & Answers

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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