Four correct numbers in the lottery

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In a lottery, $6$ distinct numbers are drawn from $1$ to $45$. A player has played $6$ numbers. What is the probability of guessing exactly $4$?

Solution

The $45$ numbers split into $6$ “winning” ones (the player’s) and $39$ “losing” ones. The draw picks $6$ numbers out of $45$.

• Possible cases: $\binom{45}{6} = 8\,145\,060$.
• Favorable cases: $4$ of the $6$ drawn are among the winning ones and the remaining $2$ among the losing ones: $\binom{6}{4}\binom{39}{2} = 15\cdot 741 = 11\,115.$

Hence $P = \frac{\binom{6}{4}\binom{39}{2}}{\binom{45}{6}} = \frac{11\,115}{8\,145\,060} \approx \boxed{0.00136}.$