InterviewSolution
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InterviewSolution
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Three unbiased coins are tossed once. What is the probability of getting (i) all head ? (ii) two heads ? (iii) one head ? (iv) at least 1 head ? (v) at least 2 heads ? |
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Answer» In tossing three coins, the sample space is given gy `S = {"HHH, HHT, HTH, THH, HTT, THT, TTH, TTT"}.` And, therefore, n(S) = 8. (i) Let `E_(1)` = event of getting all heads. Then, `E_(1) = {HHH}` and, therefore, `n(E_(1)) = 1.` `therefore` P (getting all heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/8.` (ii) Let `E_(2)` = event of getting 2 heads. Then, `E_(2) = {HHT, HTH, THH}` and, therefore, `n(E_(2)) = 3.` `therefore` P (getting 2 heads) `= P(E_(2)) = (n(E_(2)))/(n(S)) = 3/8.` (iii) Let `E_(3)` = event of getting 1 head. Then, `E_(3) = {"HTT, THT, TTH" }` and, therefore, `n(E_(3)) = 3.` `therefore` P (getting 1 head) `= P(E_(3)) = (n(E_(3)))/(n(S)) = 3/8.` (iv) Let `E_(4)` = event of getting at least 1 heads. Then, `E_(4) = {"HTT, THT, TTH, HHT, HTH, THH, HHH"}` and, therefore, `n(E_(4)) = 7.` `therefore` P (getting at least 1 head) `= P(E_(4)) = (n(E_(4)))/(n(S)) = 7/8.` (v) Let `E_(5)` = event of getting at least 2 heads. Then, `E_(5) = {"HHT, HTH, THH, HHH"}` and, therefore, `n(E_(1)) = 1.` `therefore` P (getting all heads) `= P(E_(5)) = (n(E_(5)))/(n(S)) = 4/8 = 1/2.` |
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