An urn contains four tickets marked with numbers 112, 121, 211, 222 and one ticket is drawn at random. Let Ai (i = 1, 2, 3) be the event that ith digit of the number of the ticket drawn is 1. Discuss the independence of the events A₁, A₂ and A₃.
Answers
Answer:
Step-by-step explanation:
Given that 4 tickets are numbered with 112, 121, 211, 222
A₁ be the event that 1st digit of the number is 1
P(A₁) = 2/4 = 1/2
A₂ be the event that 2nd digit of the number is 1
P(A₂) = 2/4 = 1/2
A₃ be the event that 3rd digit of the number is 1
P(A₃) = 2/4 = 1/2
A₁∩A₃ be the event that 1st digit of the number is 1 and 3rd digit
is 1
P(A₁∩A₃) = 1/4
A₁∩A₂ be the event that 1st digit of the number is 1 and 2nd digit
is 1
P(A₁∩A₂) = 1/4
A₃∩A₂ be the event that 2nd digit of the number is 1 and 3rd
digit is 1
P(A₃∩A₂) = 1/4
We can observe that
P(A₁∩A₃) = 1/4 = P(A₁)P(A₃)
P(A₁∩A₂) = 1/4 = P(A₁)P(A₂)
P(A₂∩A₃) = 1/4 = P(A₂)P(A₃)
Hence, any pair of events are independent .
Now, consider A₁∩A₂∩A₃ where all 3 digits are 1
Hence, P(A₁∩A₂∩A₃) = 0 ≠ P(A₁)P(A₂)P(A₃)
Hence,all 3 of the events are not mutually independent
together though they are pairwise independent.
Hope, it helps !
Answer:
Step-by-step explanation:Brainly.in
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dsaditya40
Secondary School Math 10+5 pts
An urn contains four tickets marked with numbers 112, 121, 211, 222 and one ticket is drawn at random. Let Ai (i = 1, 2, 3) be the event that ith digit of the number of the ticket drawn is 1. Discuss the independence of the events A₁, A₂ and A₃.
Report by Alhamd6180 24.01.2019
Answers
dsaditya40
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Answer:
Step-by-step explanation:
Given that 4 tickets are numbered with 112, 121, 211, 222
A₁ be the event that 1st digit of the number is 1
P(A₁) = 2/4 = 1/2
A₂ be the event that 2nd digit of the number is 1
P(A₂) = 2/4 = 1/2
A₃ be the event that 3rd digit of the number is 1
P(A₃) = 2/4 = 1/2
A₁∩A₃ be the event that 1st digit of the number is 1 and 3rd digit
is 1
P(A₁∩A₃) = 1/4
A₁∩A₂ be the event that 1st digit of the number is 1 and 2nd digit
is 1
P(A₁∩A₂) = 1/4
A₃∩A₂ be the event that 2nd digit of the number is 1 and 3rd
digit is 1
P(A₃∩A₂) = 1/4
We can observe that
P(A₁∩A₃) = 1/4 = P(A₁)P(A₃)
P(A₁∩A₂) = 1/4 = P(A₁)P(A₂)
P(A₂∩A₃) = 1/4 = P(A₂)P(A₃)
Hence, any pair of events are independent .
Now, consider A₁∩A₂∩A₃ where all 3 digits are 1
Hence, P(A₁∩A₂∩A₃) = 0 ≠ P(A₁)P(A₂)P(A₃)
Hence,all 3 of the events are not mutually independent
together though they are pairwise independent.
Hope, it helps !
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