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When two coins are tossed once, the sample space is given by <br> `S = {HH, HT, TH, "TT"}` and, therefore, n(S) = 4. <br> (i) Let `E_(1)` = event of getting 2 heads. Then, <br> `E_(1) = {HH}` and, therefore, `n(E_(1)) = 1`. <br> `therefore ` P (getting 2 heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/4.` <br> (ii) Let `E_(2)` = event of getting at least 1 head. Then, <br> `E_(2) = {HT, TH, HH}` and, therefore, `n(E_(2)) = 3`. <br> `therefore ` P (getting at least 1 head) `= P(E_(2)) = (n(E_(2)))/(n(S)) = 3/4.` <br> (iii) Let `E_(3)` = event of getting no head. Then, <br> `E_(3) = {"TT"}` and, therefore, `n(E_(3)) = 1`. <br> `therefore` P (getting no head) `= P(E_(3)) = (n(E_(3)))/(n(S)) = 1/4.` <br> (iv) Let `E_(4)` = event of getting 1 head or 1 tail. Then, <br> `E_(4) = {HT, TH}` and, therefore, `n(E_(4)) = 2`. <br> `therefore ` P (getting 1 head and 1 tail) `= P(E_(4)) = (n(E_(4)))/(n(S)) = 2/4 = 1/2.`