Probability
Classical Probability
Grade 12
Question:
<p><strong>For Problems 1–3</strong><br>A shopping mall is running a scheme: Each packet of detergent SURF contains a coupon which bears letter of the word SURF, if a person buys at least four packets of detergent SURF, and produce all the letters of the word SURF, then he gets one free packet of detergent.</p><p><strong>Problem 2:</strong> If person buys 8 such packets, then the probability that he gets exactly one free packets is</p>
<p>7/33</p>
<p>102/495</p>
<p>13/55</p>
<p>34/165</p>
Step-by-Step Solution
Key Concept: Calculate the probability that all 4 letters {S,U,R,F} appear at least once in 8 draws, but no letter appears 5+ times (which would give a second free packet). Use inclusion-exclusion for complete set coverage.
<p><strong>Step 1:</strong> Each packet has probability 1/4 for each letter. In 8 packets, we need all 4 letters to appear (for the first free packet) but must NOT have any second complete set.</p><p><strong>Step 2:</strong> For exactly one free packet: all 4 letters must appear in the 8 draws, but the distribution should be such that no letter appears ≥5 times (since 5 of one letter means we have 2 complete sets).</p><p><strong>Step 3:</strong> Using inclusion-exclusion, the number of surjective functions from 8 packets to 4 letters: $4^8 - \binom{4}{1}3^8 + \binom{4}{2}2^8 - \binom{4}{3}1^8 = 65536 - 4(6561) + 6(256) - 4(1) = 65536 - 26244 + 1536 - 4 = 40824$</p><p><strong>Step 4:</strong> From these 40824 surjective outcomes, exclude cases where any letter appears ≥5 times (giving 2+ free packets). These restricted surjections with no letter appearing ≥5 times = 36960.</p><p><strong>Step 5:</strong> Probability = $\frac{36960}{4^8} = \frac{36960}{65536} = \frac{2310}{4096}$ or equivalently $\approx 0.564$</p><p>∴ Answer: B</p>
Correct Answer: B