Using a Table for Probability Lesson

Adapted from Course materials (II.A Student Activity Sheet 6 & 7) for AMDM developed under the leadership of the Charles A. Dana Center, in collaboration with the Texas Association of Supervisors of Mathematics and with funding from Greater Texas Foundation.

Example 1: Driving Risks

Probability can be used to help make decisions in everyday life. Few decisions are made with absolute certainty and in most cases, the decisions we make involve a degree of risk. Sometimes we are asked to make choices with several alternatives. The choice with the highest probability of success is usually the most favorable.

Jamal will be a high school senior next year. He wants to get a vehicle to celebrate his graduation. Jamal's mother researched vehicle safety and found that 1 of every 6 teenage drivers was involved in some kind of accident. While talking to his math teacher, Jamal mentioned that he did not think the risk was high enough to be concerned about. Jamal decided to survey 500 students, 230 of whom were male, to help him convince his mother to allow him to get a vehicle. No student has both a car and a motorcycle.

Boys
	Car	Motorcycle
Boys with vehicle	150	23
Boys in accidents	40	6

Girls
	Car	Motorcycle
Girls with vehicle	225	10
Girls in accidents	15	4

Using the data, what is the probability that Jamal will be involved in an accident if he gets a motorcycle?

Of the 23 males who had a motorcycle, 6 were involved in an accident. Therefore, the probability is 26%.

6/23 = 0.26

Based on these survey data, Jamal told his mother that he only has a 1% chance of getting in an accident. Is he correct? Why or why not?

Jamal is not correct. He used the entire sample size of 500 instead of limiting his math to those students who were boys or who had vehicles. He also only used the number of accidents for motorcycles, the least number of collisions! Using these numbers, his chance of an accident is only 1.2%.

6/500 = 0.012

.012 x 100 = 1.2%

The more accurate sample to use is boys who have cars. If he does get a car, his chance of getting in an accident is closer to 27%

40/150 = .267
.267 x 100 = 26.7% or 27%

You do this one.

What is his chance of having an accident if he gets a motorcycle?

6/23 =.26 0.26 x 100 = 26%

If he gets either a car or a motorcycle, his chance of getting in an accident is 26.6%

math equation 40 + 6 ? 150 + 23 = 46/173 = 0.266

.266 x 100 = 26.6%

Example 2:

A survey of 545 winter athletes asked about their favorite snow sport.

Favorite Snow Sport
		Favorite Snow Sport
		Skiing	Snow Boarding	Tubing
Winter Sport	Cheerleading	68	41	46
	Basketball	84	56	70
	Wrestling	59	74	47

Some probabilities, we can easily determine. For instance, what is the probability of a winter athlete's favorite snow sport being skiing? Well that would be $LaTeX: P\left(ski\right)=\frac{68+84+59}{545}=\frac{211}{545}\approx0.387$

But, what if I wanted to know the probability of a person being a cheerleader, given that their favorite snow sport is skiing? Watch this video to find out:

Conditional probability of events A and B
P(B|A)=P(A and B) / P(A)

Favorite Snow Sport
		Favorite Snow Sport
		Skiing	Snow Boarding	Tubing
Winter Sport	Cheerleading	68	41	46
	Basketball	84	56	70
	Wrestling	59	74	47

Using the chart above find the following probabilities:

Determine the probability that a person prefers tubing given that the person is a basketball player.

Solution: $LaTeX: P\left(tubing\mid basketball\right)=\frac{70}{210}\approx0.333$

Determine the probability that a person is a wrestler, given that the person prefers snowboarding.

Solution: $LaTeX: P\left(wrestling\mid snowboarding\right)=\frac{74}{171}\approx0.433$

Determine the probability that a person prefers skiing, given that the person is a cheerleader.

Solution: $LaTeX: P\left(skiing\mid cheerleader\right)=\frac{68}{155}\approx0.439$

Let's say I wanted to determine the probability of a person being a cheerleader or preferring skiing. Let's discuss how we might determine that probability.

Probability of events, A or B
P(A or B)=P(A)+P(B)-P(A and B)

Let's apply that rule to this example: Let's say you have 10 kittens, 4 are males and 6 are females. 3 males are gray, and 2 females are gray, the rest are black. Determine the probability of randomly selecting a kitten that is gray or female.

$P\left(gray\:or\:female\right)=P\left(gray\right)+P\left(female\right)-P\left(gray\:and\:female\right)=\frac{5}{10}+\frac{6}{10}-\frac{2}{10}=\frac{9}{10}=0.9$

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