First prize token on the fourth draw

Question

A box holds 3 prize tokens and 7 blank tokens, 10 in total. You draw tokens one at a time, without replacement. What is the probability that the first prize token appears on exactly the fourth draw?

Accepted Answer

"First prize on the fourth draw" pins down the earlier draws too: the first three must all be blanks, and the fourth must be a prize. Chain the draws together, multiplying and shrinking the box as you go. Draws 1 to 3 all blank. Blanks start at 7 of 10, then 6 of 9, then 5 of 8: \frac{7}{10}	imes\frac{6}{9}	imes\frac{5}{8}. Draw 4 is a prize. After 3 blanks are gone, 3 prizes remain out of the 7 tokens left: \frac{3}{7}. Multiply everything: \frac{7}{10}	imes\frac{6}{9}	imes\frac{5}{8}	imes\frac{3}{7}=\frac{630}{5040}=\frac{1}{8}=0.125. Takeaway. "First success on draw k" means "the firs…

First prize token on the fourth draw

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