ProbabilityHard
Question
A bag contains four tickets marked with numbers 112, 121, 211, 222. One ticket is drawn at random from the bag. Let Ei(i = 1, 2, 3) denote the event that ith digit on the ticket is 2. Then
Options
A.E1 and E2 are independent
B.E2 and E3 are independent
C.E3 and E1 are independent
D.E1, E2, E3 are independent
Solution
E1 → p first digit is ′2′ ⇒ 211 or 222
P(E1) =
E2 → second digit is ′2′ ⇒ 121, 222
P (E2) =
E3 third digit is ′2′ ⇒ 222 & 112
P(E3) =
(E1 ∩ E2) = 222 ⇒ P(E1 ∩ E2) =
= P (E1) P (E2)
Similiarly E2 & E3, E1 & E3
also E1 ∩ E2 ∩ E3 → 222 ⇒ P(E1 ⎳ E2 ∩ E3) =
P(E1) P(E2) P(E3)
P(E1) =
E2 → second digit is ′2′ ⇒ 121, 222
P (E2) =
E3 third digit is ′2′ ⇒ 222 & 112
P(E3) =
(E1 ∩ E2) = 222 ⇒ P(E1 ∩ E2) =
= P (E1) P (E2)Similiarly E2 & E3, E1 & E3
also E1 ∩ E2 ∩ E3 → 222 ⇒ P(E1 ⎳ E2 ∩ E3) =
P(E1) P(E2) P(E3)Create a free account to view solution
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