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riadik2000 [5.3K]
3 years ago
8

Roger will flip a coin 3 times. Which 2 events are disjoint?

Mathematics
1 answer:
Ganezh [65]3 years ago
7 0

Answer: D. Getting at least 2 heads and Getting at least 2 tails

=========================================================

Explanation:

Choice A is not disjoint because "getting at least 1 tail" could have the sequence TTH. We see that we have two tails and exactly one head. So the events "getting exactly 1 head" and "Getting at least 1 tail" are possible to occur at the same time; therefore, they aren't disjoint events. Disjoint events are two events that cannot occur simultaneously. An example would be flipping heads and tails on the same coin at the same time.

Choice B is also not disjoint. We could have the sequence THT, which has at least one head and at least one tail. "At least" means that amount or more.

Choice C is also not disjoint. We could have the sequence HTT. This has exactly one head and at least two tails.

Choice D is the only thing left. It must be the answer. It is not possible to get 2 heads and 2 tails when Roger only flips the coin 3 times. He would need to flip the coin at least 4 times for this to happen. The portions "at least" don't even need to be considered. So this shows how choice D is disjoint.

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<em>The table that shows the relative frequency of data is in the below further explanation.</em>

\texttt{ }

<h3>Further explanation</h3>

A set is a clearly defined collection of objects.

To declare a set can be done in various ways such as:

  • With words or the nature of membership
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\texttt{ }

Multiplying set A x B is by pairing each member of set A with each member of set B.

<u>Example:</u>

<em>A = {1, 2, 3}</em>

<em>B = {a, b}</em>

Then

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\texttt{ }

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Intersection of set A and B ( A ∩ B ) is to find the members that are both in Set A and Set B.

<u>Example:</u>

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Let us now tackle the problem!

\texttt{ }

<em>A sports club has 84 members who learned baseball, and 42 of those members also learned basketball.</em>

\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&\boxed{42}& \\Don't Play Basketball& & \end{array}\right]

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Number of members who learned baseball but did not learn basketball will be ( 84 - 42 ) = 42 members

\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&42& \\Don't Play Basketball&\boxed{42}& \end{array}\right]

\texttt{ }

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\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&42&\boxed{25} \\Don't Play Basketball&42& \end{array}\right]

\texttt{ }

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\left[\begin{array}{ccc}&Play Baseball&Don't Play Baseball\\Play Basketball&42&25 \\Don't Play Basketball&42&\boxed{8} \end{array}\right]

\texttt{ }

<em>Total Number of Students = 42 + 42 + 25 + 8 = 117</em>

\texttt{ }

Table of relative frequency

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\texttt{ }

<h3>Learn more</h3>
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<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Sets

Keywords: Sets , Venn , Diagram , Intersection , Union , Mean , Median , Mode

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