$\frac{1}{2}$

$\frac{1}{3}$

$\frac{3}{2}$

$\frac{3}{4}$

D. $\frac{3}{4}$

Total 4 cases = [HH, TT, TH, HT]

Favourable cases = [HH, TH, HT]

Please note we need atmost one tail, not atleast one tail.

So probability = $\frac{3}{4}$

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{52}{55}$

$\frac{3}{55}$

$\frac{41}{44}$

$\frac{3}{44}$

$\frac{3}{4}$

$\frac{1}{4}$

$\frac{7}{4}$

$\frac{1}{2}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{3}{2}$

$\frac{3}{4}$

$\frac{1}{3}$

$\frac{1}{9}$

$\frac{1}{12}$

$\frac{2}{9}$

$\frac{2}{3}$

$\frac{8}{21}$

$\frac{3}{7}$

$\frac{9}{22}$

$\frac{7}{19}$

$\frac{6}{19}$

$\frac{5}{19}$

$\frac{4}{19}$

$\frac{4}{13}$

$\frac{1}{52}$

$\frac{1}{4}$

None of above

1

2

$\frac{1}{2}$

0

30%

35%

40%

45%

$\frac{2}{121}$

$\frac{2}{221}$

$\frac{1}{221}$

$\frac{1}{13}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{5}$

$\frac{1}{6}$

1

2

$\frac{1}{2}$

0

$\frac{1}{13}$

$\frac{2}{13}$

$\frac{1}{26}$

$\frac{1}{52}$

1

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{4}{3}$