Probability
Drawing without Replacement
Grade 12
Question:
<p>A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the seventh draw is:</p>
<p>(a) \(\frac{15}{286}\)</p>
<p>(b) \(\frac{105}{286}\)</p>
<p>(c) \(\frac{35}{286}\)</p>
<p>(d) \(\frac{7}{286}\)</p>
Step-by-Step Solution
Key Concept: The last ball drawn must be a black ball, and exactly 2 of the remaining 3 black balls must appear in the first 6 draws.
<p>For the procedure to end at the seventh draw, the 7th ball must be black (the third black ball) and exactly 2 black balls must be among the first 6 balls drawn.</p><p>This means: 4 white balls and 2 black balls in first 6 draws, then 1 black ball on 7th draw.</p><p>Probability = \(\frac{\binom{10}{4}\binom{3}{2}}{\binom{13}{6}} \times \frac{1}{7} = \frac{15}{286}\)</p>
Correct Answer: A