Question

There are 2020 amoebas in a room undergoing reverse mitosis. Every second, two amoebas are chosen uniformly at random and combined to form one single amoeba. Find the probability that after 2018 seconds there is some amoeba that has never been combined with another amoeba.

Answer 1

Let us name ameba Page Nol = 2020 Number of amoebas at the intial stage Every seconds two amoebas are combined Perom one single amoeba. Find the probability that after 2018 seconds there is some amoeba that has never combined with another amoeba. Sol amoebas as 2020 Let Ei be event that after 2018 seconds number i is not combined with mother amoeba. Then PLE,)=P(E). P(E2020) If E is the probability that some amoeba is not combined with another ameba after 2018 seconds then PLE) PLE, ao Es of E2020 PCE) 2020 PCE,) Note There cannot be two amoeba's which are not combined with another 2018 seconds. P (Amoeba Number I is not combined with another amoeba Here only remaining 2019 ambebas are combined ) P(Amoeba No. 1 is not combined with another amoeba after two. seconds) Es-20) - amoeba in after Isecond) 20195 ( 2020 2C2 .se 2019 (2 20202 201802 X 2019,

Note Number of amoebas left over after Isec, 3 sec, 2018 sec are 2019, 2018, 2017 2 sec. / is not combined with another P (Amoeba Nol amoeba after 3 seconds) 2019 20186 20172 X X 2020, 2019 (2 Contio inning in this way 2018 201842 2019 PLE,) = 2017 х ner/ 2020 x Х 02C 2019 nir cancellation take place 222 i. P(E) 202% 1 x 2 2020x2019 2 .. PLE) = 2020 PLE) - 2020 X 2020X2019 ) PLE)= 2019