<p>A man throws a fair coin a number of times and gets 2 points for each head he throws and 1 point for each tail he throws. The probability that he gets exactly 6 points is</p>
<p>(a) \(\frac{21}{32}\)</p>
<p>(b) \(\frac{23}{32}\)</p>
<p>(c) \(\frac{41}{64}\)</p>
<p>(d) \(\frac{43}{64}\)</p>
Step-by-Step Solution
Key Concept: Find all combinations of heads and tails that sum to exactly 6 points, calculate probability for each, and sum them.
<p><strong>Step 1:</strong> To get exactly 6 points, the man can have: 3 heads and 0 tails (3×2=6) or 2 heads and 2 tails (2×2+2×1=6) or 1 head and 4 tails (1×2+4×1=6) or 0 heads and 6 tails (0×2+6×1=6).</p><p><strong>Step 2:</strong> Required combinations: (3H,0T), (2H,2T), (1H,4T), (0H,6T).</p><p><strong>Step 3:</strong> P(3H) = $\binom{3}{3}\left(\frac{1}{2}\right)^3 = \frac{1}{8}$; P(2H,2T) = $\binom{4}{2}\left(\frac{1}{2}\right)^4 = \frac{6}{16}$; P(1H,4T) = $\binom{5}{1}\left(\frac{1}{2}\right)^5 = \frac{5}{32}$; P(0H,6T) = $\left(\frac{1}{2}\right)^6 = \frac{1}{64}$</p><p><strong>Step 4:</strong> Total = $\frac{1}{8} + \frac{6}{16} + \frac{5}{32} + \frac{1}{64} = \frac{8+24+10+1}{64} = \frac{43}{64}$</p><p>∴ Answer is (c).</p>
Correct Answer: A