Raffle

There are 200 draws in the raffle, but only 20 of them win. What is the probability of at least 4 winnings for a group of people who have bought 5 tickets together?

Correct result:

p =  30.78712 %

Solution:

$$\begin{gathered} C_5(20)=\left({\displaystyle \genfrac{}{}{0px}{}{20}{5}}\right)={\displaystyle \frac{20!}{5!(20-5)!}}={\displaystyle \frac{20\cdot 19\cdot 18\cdot 17\cdot 16}{5\cdot 4\cdot 3\cdot 2\cdot 1}}=15504 \\ n=\left(\genfrac{}{}{0px}{}{20}{5}\right)=15504 \\ {a}_1=20\cdot 19\cdot 18\cdot 17\cdot 180\cdot 179\cdot 178\cdot 177\cdot 176\cdot 175\cdot 174\cdot 173\cdot 172\cdot 171\cdot 170\cdot 169\cdot 168\cdot 167\cdot 166\cdot 165=7.10637079329\cdot 10^{40} \\ {b}_1=200\cdot 199\cdot 198\cdot 197\cdot 196\cdot 195\cdot 194\cdot 193\cdot 192\cdot 191\cdot 190\cdot 189\cdot 188\cdot 187\cdot 186\cdot 185\cdot 184\cdot 183\cdot 182\cdot 181=3.92570096972\cdot 10^{45} \\ {a}_2=20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 180\cdot 179\cdot 178\cdot 177\cdot 176\cdot 175\cdot 174\cdot 173\cdot 172\cdot 171\cdot 170\cdot 169\cdot 168\cdot 167\cdot 166=6.8910262238\cdot 10^{39} \\ {p}_1=100\cdot n\cdot {\displaystyle \frac{a_1}{b_1}}=100\cdot 15504\cdot {\displaystyle \frac{7.10637079329\cdot 10^{40}}{3.92570096972\cdot 10^{45}}}\doteq 28.0656\mathrm{\%} \\ {p}_2=100\cdot n\cdot {\displaystyle \frac{a_2}{b_1}}=100\cdot 15504\cdot {\displaystyle \frac{6.8910262238\cdot 10^{39}}{3.92570096972\cdot 10^{45}}}\doteq 2.7215\mathrm{\%} \\ p=p_1+p_2=28.0656+2.7215=30.78712\mathrm{\%} \end{gathered}$$

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Math student

it rly helped

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