Check sibling questions

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

Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 2
Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 3
Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 4
Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 5
Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 6
Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 7
Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 8
Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 9
Ex 16.3, 8 - Chapter 16 Class 11 Probability - Part 10

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Transcript

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 Three coins are tossed once. Find the probability of getting (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 Three coins are tossed once. Find the probability of getting (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 Three coins are tossed once. Find the probability of getting (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 Three coins are tossed once. Find the probability of getting (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 Three coins are tossed once. Find the probability of getting (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 Three coins are tossed once. Find the probability of getting (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 Three coins are tossed once. Find the probability of getting (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 Three coins are tossed once. Find the probability of getting (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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.