Problem statement: There are 5 pair of shoes in a shoe rack. Four shoes are drawn one by one at random. Find the probability that at least one pair of shoes is drawn.
I proceeded as follows:
Method 1: First shoes can be drawn in 10 ways, 2nd in 9 ways,…, 4th in 7 ways. Hence, total no. of ways of drawing a shoes =$10.9.8.7$
Case 1: Only 1 pair is drawn.
Out of 5 pairs, 1 can be chosen in $^5C_1=5$ ways and then rest of the two shoes are to be selected from rest of the 8 shoes such that we don't draw any pair. We can select the third shoe out of balance 8 in $8$ ways and fourth in $6$ ways. Hence, favourable no of ways =$5.8.6$
Case 2: Two pairs are drawn.
It can be done in $^5C_2 \times 2=20$ ways. First shoes can be any of the 10 shoes,
Hence the required probability =$\frac{20+5.8.6}{10.9.8.7}=13/(252)$
Method 2: Out of $10$ shoes, 4 can be selected in $^{10}C_4$ ways.
Case 1: Only one pair is selected. Same as above.
Case 2: Same as above.
Required probability =$\frac{20+5.8.6}{^{10}C_4}=27/21\gt 1$, which is obviously wrong.
Method 3: Required probability =$1-$probability that no pair is drawn. =$1-(10/10)(8/9)(6/8)(4/7)=13/21$.
I think I have exhausted all the cases. I don't understand why answers are different from these different methods. What am I not seeing?
Please help me understand. Thanks.
Best Answer
Choosing no match means you choose exactly one shoe from each of four original pairs, meaning there is one pair untouched. There are five original pairs that can be so untouched. Given this condition, you are free to choose either of the two shoes in each of the four pairs you do selected from.
Hence the number of "non-matching" ways is $5 \cdot 2^4$. Divide this by the number of ways to choose $4$ shoes out of the original $10$ without conditions, i.e., ${10 \choose 4}$.