Question

Seth is analyzing the number of students in his class and high school who own skateboards. He puts
the data in the table shown. Given the information in the table, which event is the least likely?

Answer 1

Answer:it is the third option

Step-by-step explanation:

I took the test (:

Answer 2

Least likely event means probability of its occurrence is minimum among other events. The least likely events from the given events is: Option B: A randomly selected student who is a freshmen owns a skateboard

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

What is chain rule in probability?

For two events A and B, by chain rule, we have:

$P(A \cap B) = P(B)P(A|B) = P(A)P(A|B)$

How to find the conditional probability?

Suppose that there are two events A and B. Then suppose the conditional probability are:

• P(A|B) = probability of occurrence of A given B has already occurred.
• P(B|A) = probability of occurrence of B given A has already occurred.

We can then use the chain rule to find them, or Baye's theorem also helps in finding these probabilities.

We are given the following table of joint relative frequency:

$\begin{array}{cccc}&\text{Freshmen(B)}&\text{Non-Freshmen(B')}&\text{Total}\\\text{Owns a skateboard(A)}&40&110&150\\\text{Doesn't own skateboard(A')}&210&840&1050\\\text{Total}&250&950&1200\end{array}$

We named events as A, A' and B and B' (event and their complements).

The important probabilities which are going to be used are evaluated as:
$P(A) = \dfrac{n(A)}{n(S)} = \dfrac{150}{1200} = \dfrac{1}{8}$

$P(B) = \dfrac{n(B)}{n(S)} = \dfrac{250}{1200} = \dfrac{5}{24}$

$P(A') = 1 - P(A) = \dfrac{7}{8}\\\\P(B') = 1-P(B) = \dfrac{5}{24} = \dfrac{19}{24}$

$P(A \cap B) = \dfrac{n(A \cap B)}{n(S)} = \dfrac{40}{1200} = \dfrac{1}{30}\\\\P(A' \cap B') = \dfrac{n(A' \cap B')}{n(S)} = \dfrac{840}{1200} = \dfrac{7}{10}\\P(A' \cap B) = \dfrac{n(A' \cap B)}{n(S)} = \dfrac{210}{1200} = \dfrac{7}{40}$

Evaluating probabilities of choices given, we get:

Case 1: E = A randomly selected student who owns a skateboard is a freshmen

P(E) = ?

P(E) = P(B|A) = $\dfrac{P(A \cap B)}{P(A)} = \dfrac{1/30}{1/8} = \dfrac{4}{15}$

Case 2: E = A randomly selected student who is a freshmen owns a skateboard:

P(E) = P(A|B) = $\dfrac{P(A \cap B)}{P(B)} = \dfrac{1/30}{5/24} = \dfrac{4}{25}$

Case 3: E = A randomly selected student who doesn't owns a skateboard is a freshmen

P(E) = P(B|A') = $\dfrac{P(A' \cap B)}{P(A')} = \dfrac{7/40}{19/24} = \dfrac{21}{95}$

Case 4: E = A randomly selected student who doesn't owns a skateboard is not a freshmen

P(E) = P(B'|A') = $\dfrac{P(A' \cap B')}{P(A')} = \dfrac{7/10}{19/24} = \dfrac{84}{95}$

Thus, the least probability is for the second case.

Thus, the least likely event(the event having least probability) from the given events is: Option B: A randomly selected student who is a freshmen owns a skateboard

Learn more about probability here:

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