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3 years ago

Which of the following are independent pairs of events? Check all that apply.

Mathematics

2 answers:

Mila [183]3 years ago

4 0

Answer:

A,D and E

Step-by-step explanation:

Intuitively speaking two events are independents when the occurrence of one event (A) does not exert influence over the event (B)

Symbolic speaking in terms of sets

P(A∩B)= P(A) * P(B)

Examining each alternative

A. Flipping two coins and having one show a head and the other show a head

Independent. Each flipping coin does not exert any influence on the other

B. Being sneezed on and catching a cold

Dependent. This sequence of events expresses cause consequences effects.

C. When rolling a pair of dice, getting a total of 6 and having one die show a 3

Dependent

Under these conditions, the second event is dependent since all we need is a sum of six, having the first dice showing three. The second one has to show three. So it does exert influence under the result.

D. Rolling a 4 on one die and rolling a 2 on another die

Independent. To get a 4 does not exert any influence on getting a 2 on another dice.

E. Seeing clouds and winning the lottery

Independent.

There is no influence seeing clouds and winning the lottery.

Ulleksa [173]3 years ago

4 0

Independent events are exclusive or non-mutual events that when one event happens, its occurrence does not affect the occurrence of the other event. in this case, A could be a case. Also, options D and E could be another case of independent events

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Step-by-step explanation:

-14x > -112

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-14x/-14 < -112/-14

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Complete Question

The complete question is shown on the first uploaded image

Answer:

$P(U |D ) = 0.198$

$P(O\ n \ B) = 0.188$

$P(O | B) = 0.498$

Step-by-step explanation:

The total number of deaths is mathematically represented as

$T = 16 + 23 + \cdots + 16$

$T =623$

The total number of deaths in 1996 - 2000 is mathematically represented as

$T_a = 16 + 23+ \cdots + 30$

$T_a = 235$

The total number of deaths in 2001 - 2005 is mathematically represented as

$T_b = 17 + 16 + \cdots + 23$

$T_b = 206$

The total number of deaths in 2006 - 2010 is mathematically represented as

$T_c = 15 + 17 + \cdots + 16$

$T_d = 182$

Generally the the probability that it would occur under the tree given that the death was after 2000 is mathematically represented as

$P(U |D ) = \frac{P(A \ n\ U )}{P(A)}$

Here $P(A \ n\ U )$ represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

$P(A \ n\ U ) = \frac{Z}{ T}$

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

$Z = 35 + 42$

=> $Z = 77$

=> $P(A \ n\ U ) = \frac{77}{ 623}$

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$P(A)$ is the probability of the death occurring after 2000 and this is mathematically represented as

$P(A) = \frac{T_b + T_c}{ T}$

=> $P(A) = \frac{ 206+ 182}{623}$

$P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+ 182}{623}}$

=> $P(U |D ) = 0.198$

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

$P(O | B) = \frac{T_z}{ T}$

Here $T_z$ is the total number of death outside / camping before 2001 and the value is 117

$P(O \ n \ B) = \frac{117}{623}$

=> $P(O\ n \ B) = 0.188$

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

$P(O | B) = \frac{ P( O \ n \ B)}{ P(B)}$

Here $P(B)$ is the probability that it was before 2001 , this is mathematically represented as

$P(B ) = \frac{T_a}{T}$

=> $P(B ) = \frac{235}{623}$

$P(O | B) = \frac{ \frac{117}{623}}{ \frac{235}{623}}$

=> $P(O | B) = 0.498$

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