Theoretical and Experimental Probability
Lesson 2-6
Algebra 1
Additional Examples
A bowl contains 12 slips of paper, each with a different name of a
month on it. Find the theoretical probability that a slip selected at random
from the bowl has the name of a month that ends with “ber.”
number of favorable outcomes
P(event) = number of possible outcomes
4
= 12
1
= 3
There are 4 months out of 12 that end with “ber”:
September, October, November, and December
Simplify.
1
The probability of picking a month that ends with “ber” is 3 .
Theoretical and Experimental Probability
Lesson 2-6
Algebra 1
Additional Examples
For a number cube, find the probability of not rolling a
number divisible by 3.
number of favorable outcomes
2
1
P(÷ 3) = number of possible outcomes = 6 = 3
P(not ÷ 3) = 1 – P(÷ 3)
1
2
=1– 3 = 3
Use the complement formula.
Simplify.
2
The probability of not rolling a number divisible by 3 is 3 .
Theoretical and Experimental Probability
Lesson 2-6
Additional Examples
Algebra 1
Find the odds, for the spinner below, in favor of the spinner
landing on an even number.
Favorable outcomes: 2, 4, 6, 8
Unfavorable outcomes: 1, 3, 5, 7
The odds are 1 : 1 in favor of the event.
Theoretical and Experimental Probability
Lesson 2-6
Algebra 1
Additional Examples
Quality control inspected 500 belts at random. They found no
defects in 485 belts. What is the probability that a belt selected at
random will pass quality control?
number of times an event occurs
.
P(no defects) = number of times the experiment is done
485
= 500
Substitute.
= 0.97 = 97%
Simplify. Write as a percent.
The probability that a belt has no defects is 97%.
Theoretical and Experimental Probability
Lesson 2-6
Algebra 1
Additional Examples
If the belt manufacturer from Additional Example 4 has 6258
belts, predict how many belts are likely to have no defects.
number with no defects = P(no defects) • number of belts
= 0.97 • 6258
Substitute. Use 0.97 for 97%.
= 6070.26
Simplify.
Approximately 6070 belts are likely to have no defects.