Ex 16.3, 8 - Three coins are tossed once. Find probability - Basic Formula

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  1. Chapter 16 Class 11 Probability
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Ex 16.3, 8 Three coins are tossed once. Find the probability of getting • 3 heads If 3 coins are tossed various combination possible are S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} n(S) = 23= 8 Let A be the event of getting 3 head A = {HHH} n(A) = 1 Probability of 3 heads = P(A) = n(A)﷮n(S)﷯ = 𝟏﷮𝟖﷯ Ex 16.3, 8 (ii) 2 heads Let B be the event of getting 2 head B = {HHT, HTH, THH} n(B) = 3 Probability of getting 2 head = P(B) = n(B)﷮n(s)﷯ = 𝟑﷮𝟖﷯ Ex 16.3, 8 (iii) at least 2 heads Let C be the event of getting at least 2 heads. C = {HHH, HHT, HTH, THH} n(C) = 4 Probability of getting at least 2 heads = P(C) = n(C)﷮n(s)﷯ = 4﷮8﷯ = 𝟏﷮𝟐﷯ Ex 16.3, 8 (iv) at most 2 heads Let D be the event of getting at most 2 heads. i.e. getting 0 head, 1 head or 2 head D = {HHT, HTH, THH, HTT, THT, TTH, TTT} So, n(D) = 7 Probability of getting at most 2 heads = P(D) = n(D)﷮n(s)﷯ = 𝟕﷮𝟖﷯ Ex 16.3, 8 (v) no head No head means all tails are obtained Let E be the event of getting no head So, E = {TTT} n(E) = 1 Probability of getting no head = P(E) = n(E)﷮n(s)﷯ = 𝟏﷮𝟖﷯ Ex 16.3, 8 (vi) 3 tails Let F be the event of getting 3 tails. F = {TTT} n(F) = 1 Probability of getting 3 tails = P(F) = n(F)﷮n(s)﷯ = 𝟏﷮𝟖﷯ Ex 16.3, 8 (vii) exactly two tails Let G be the event of getting exactly 2 tails G = {HTT, THT, TTH} n(G) = 3 Probability of getting exactly two tails = P(G) = n(G)﷮n(s)﷯ = 𝟑﷮𝟖﷯ Ex 16.3, 8 (viii) no tail Let H be the event of getting no tail H = {HHH} n(H) = 1 Probability of getting no tails = P(H) = n(H)﷮n(s)﷯ = 𝟏﷮𝟖﷯ Ex 16.3, 8 (ix) at most two tails. Let I be the event of getting at most 2 tails. i.e. getting 0 tails, 1 tail or 2 tail I = {HHH, HHT, HTH, THH, HTT, THT, TTH} n(I) = 7 Probability of getting at most two tails = P(I) = n(I)﷮n(s)﷯ = 𝟕﷮𝟖﷯

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