11. One Match (from 8th grade, 2 points). In the lottery ticket "6 out of 45," there are a total of 45 natural numbers. From these 45 numbers, a lottery participant must choose a combination that contains exactly 6 numbers. Two participants independently choose their combinations. What is the probability that their combinations will have exactly one common number?
Solution. Let's mark with some secret sign the six numbers chosen by the first participant. This results in 6 marked numbers and 39 unmarked numbers. We need to find the probability that the combination chosen by the second participant includes exactly 1 marked number and exactly 5 unmarked numbers. The probability of this is $\frac{C_{6}^{1} C_{39}^{5}}{C_{45}^{6}} \approx 0.424$. Answer: approximately 0.424.