\(\frac{1}{4}\)
\(\frac{1}{2}\)
1
\(\frac{43}{78}\)
\(\frac{3}{4}\)
Correct answer is E
Prob. of getting at least one head Prob. of getting one head + prob. of getting 2 heads = \(\frac{1}{4}\) + \(\frac{2}{4}\) = \(\frac{3}{4}\)
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