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\[ A = P(1 + \fracrn)^nt \] - Get link 4share

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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\frac{4!}{2!1!1!} = 12
So total favorable sequences:
$4 \times 3 \times 12 = 144$
Therefore, the desired probability is:
\frac{144}{256} = \frac{9}{16}
Wait — but this counts only sequences with one pair and two singles, a...
But note: the problem says exactly two of the chosen options are the s...
Thus, our count is correct.
Hence, the probability is:
But wait: $144 \div 16 = 9$, $256 \div 16 = 16$, yes: $\frac{144}{256}...

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