Question

If A and B are mutually exclusive events, then LaTeX: P\left(A\cap B\right)=0 P ( A ∩ B ) = 0.

Group of answer choices

True

False

A pair of dice is tossed. Events A and B are defined as follows.

A:{ The sum of numbers on the dice is 2}

B:{ At least one of the dice shows a "2"}

Then LaTeX: P\left(A\cap B\right)=0 P ( A ∩ B ) = 0.

Group of answer choices

False

True

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Question 33 pts

Find LaTeX: \binom{6}{3}. ( 6 3 ) .

Group of answer choices

30

20

24

120

Two events are listed below:

A: {The home's roof is less than 10 years}

B: {The home has a security system}

Define the event LaTeX: A^c\cup B A c ∪ B

Group of answer choices

The home's roof is at least 10 years or it has a no security system

The home's roof is at least 10 years and it has a security system

The home's roof is less than 10 years or it has a security system

The home's roof is at least 10 years or it has a security system

Answer 1

Answer:

True

True

20

The home's roof is at least 10 years or it has a security system

Step-by-step explanation:

The two events are mutually exclusive when they have no common outcome i.e. A∩B={ }, so, P(A∩B)=0. Thus, the statement is true.

When a pair of dice is thrown and event A is sum of number is 2 and event B is at least one dice show 2. So,

A={(1,1) and B={(1,2),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2)}

A∩B={ }, So, P(A∩B)=0. So, the statement is true.

We have to find 6C3. We know that nCx=n!/x!(n-x)!.

6C3=6!/3!(6-3)!

6C3=6!/3!3!

6C3=6*5*4*3!/3!3!

6C3=120/6=20

For events A={The home's roof is less than 10 years}  and B={The home has a security system} the event (A^c)∪B is describes as The home's roof is not less than 10 years or the home has a security system.It means that (A^c)∪B describe that the home's roof is at least 10 years or it has a security system

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