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  1. Chapter 13 Class 12 Probability
  2. Serial order wise

Transcript

Example 12 Three coins are tossed simultaneously. Consider the event E ‘three heads or three tails’, F ‘at least two heads’ and G ‘at most two heads’. of the pairs (E,F), (E,G) & (F,G), which are independent? which are dependent? Two events A and B are independent if P(A ∩ B) = P(A) . P(B) Three coins are tossed simultaneously S = {(H, H, H), (H, H, T), (T, H, H), (H, T, H), (T, T, H), (T, H, T), (H, T, T), (T, T, T)} Let us define 3 events as E : 3 head or 3 tails F : atleast two heads G : atmost two heads E : 3 head or 3 tails E : {HHH, TTT} P(E) = 2/8 = 1/4 vF : atleast two heads F : {HHH, HHT, HTH, THH} P(F) = 4/8 = 1/2 G : atmost two heads G : {HHT, HTH, THH HTT, THT, TTH ,TTT } P(G) = 7/8 Finding probabilities of E, F and G Now, let us find Probabilities of E ∩ F , F ∩ G , E ∩ G E ∩ F = 3 head = {HHH} So, P(E ∩ F) = 1/8 Now, P(E) . P(F) = 1/4 × 1/2 = 1/8 P(E ∩ F) = P(E).P(F) Thus, E & F are independent events F ∩ G = Two head = {HHH, HTH, THH} So, P(F ∩ G) = 3/8 Now, P(F) . P(G) = 1/2 × 7/8 = 7/16 P (F ∩ G) ≠ P(F) . P(G) Thus, F & G are not independent events E ∩ G = 3 tails = {TTT} So, P(E ∩ G) = 1/8 Now, P(E) . P(G) = 1/4 × 7/8 = 7/32 P (E ∩𝐆) ≠ P (E). P(G) Thus, E & G are not independent events

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.