Respuesta :

Kalahira
If two events are independent, the occurrence of one event does not affect the other. That is if two events are independent, then P(Aâ©B)=P(A)P(B) Let A be the even getting a sum of 6 in a single roll of two dice. Sample space of A ={(1,5)(5,1)(2,4)(4,2)(3,3)} n(A)=5; n(S)=36 Therefore P(A) =n(A)/n(S) =5/36 ---------(1) Let B be the event of rolling doubles. Sample space for B ={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)} n(B)=6;n(S)=36 P(B) = n(B)/n(S) = 6/36 --------------(2) Aâ©B is the event of getting a sum of 6 and rolling doubles. Therefore n(Aâ©B)=1 P(Aâ©B)=1/36 ------(3) Multiplying equation (1) and (2) (5/36)*(6/36)=5/216 but P(Aâ©B)=1/36 P(Aâ©B) ≠P(A)P(B) Therefore, the events are not independent.

Answer:

The events, rolling a sum of 6 and rolling doubles are not  independent events

Step-by-step explanation:

We know that two events A and B are independent if:

[tex]P(A\bigcap B)=P(A)\times P(B)[/tex]

where P denotes the probability of an event.

Let A denote the event of rolling a sum of 6

and B denote the event of rolling a double.

Then A∩B denote the event of rolling a double whose sum is 6.

Now we know that there are a total of 36 outcomes on rolling two die

Now [tex]P(A)=\dfrac{5}{36}[/tex]

( Since there are 5 outcomes such that sum of the roll is: 6

i.e. (1,5), (2,4) , (3,3) , (4,2) and (5,1) )

Also,

[tex]P(B)=\dfrac{6}{36}=\dfrac{1}{6}[/tex]

(    Since there are a total of 6 events which are double

i.e.  (1,1) (2,2) (3,3) (4,4) (5,5) and (6,6)  )

This means that:

[tex]P(A)\times P(B)=\dfrac{5}{36}\times \dfrac{1}{6}\\\\\\P(A)\times P(B)=\dfrac{5}{216}[/tex]

Also, A∩B is the outcome (3,3)

Hence, we have:

 [tex]P(A\bigcap B)=\dfrac{1}{36}[/tex]

Hence, we get:

[tex]P(A\bigcap B)\neq P(A)\times P(B)[/tex]

         Hence, the events are not independent.

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