# Example 12 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

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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Chapter 13 Class 12 Probability (Term 2)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.