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NCERT Class XI Mathematics - Probability - Solutions

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Question : 31 of 54
Marks: +1, -0
Three coins are tossed once. Find the probability of getting
(i) 3 heads
(ii) 2 heads
(iii) atleast 2 heads
(iv) atmost 2 heads
(v) no head
(vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) atmost two tails
Solution:  
An experiment consists of tossing 3 coins
∴ The sample space of the given experiment is given by
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
∴ n(S) = 8
(i) Let E be the event that 3 heads appear
∴ n(E) = 1 (Since E = {HHH})
⇒ P (E) = n(E)n(S)\frac{n(E)}{n(S)} = 18\frac{1}{8}
(ii) Let E be the event that 2 heads appear
∴ n(E) = 3 (Since E = {HHT, HTH, THH})
⇒ P (E) = n(E)n(S)\frac{n(E)}{n(S)} = 38\frac{3}{8}
(iii) Let E be the event that atleast 2 heads appear
∴ n(E) = 4 (Since E = {HTH, HHT, THH, HHH})
⇒ P (E) = n(E)n(S)\frac{n(E)}{n(S)} = 48\frac{4}{8} = 12\frac{1}{2}
(iv) Let E be the event that at most 2 heads appear
∴ n(E) = 7 [Since E = HHT, HTH, THH, TTT, THT, TTH, HTT]
⇒ P (E) = n(E)n(S)\frac{n(E)}{n(S)} = 78\frac{7}{8}
(v) Let E be the event that no head appears
∴ n(E) = 1 (Since E = {TTT}) ⇒ P (E) = n(E)n(S)\frac{n(E)}{n(S)} = 18\frac{1}{8}
(vi) Let E be the event that 3 tails appear
∴ n(E) = 1 (Since E = {TTT})
⇒ P (E) = n(E)n(S)\frac{n(E)}{n(S)} = 18\frac{1}{8}
(vii) Let E be the event that exactly two tails appear
∴ n(E) = 3 (Since E = {TTH, THT, HTT})
⇒ P (E) = n(E)n(S)\frac{n(E)}{n(S)} = 38\frac{3}{8}
(viii) Let E be the event that no tail appears
∴ n(E) = 1 (Since E = {HHH})
(ix) Let E be the event that atmost two tails appear
∴ n(E) = 7 (Since E = {THH, HTH, HHT, HTT, THT, TTH, HHH})
⇒ P (E) = n(E)n(S)\frac{n(E)}{n(S)} = 78\frac{7}{8}
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