Question

Question 5 6 marks A Japanese avocado tree has twenty fruits, exactly fifteen of which have no stones. A monkey eats five whole avocados, picked at random. Subsequently, an avocado is picked randomly from the remaining fifteen 2(a) What is the probability that this avocado contains a stone? b) Given that this avocado contains a stone, what is the probability that the monkey consumed at least one stone?

Answer 1

Given data no. of fruits =20 no. of fruits that have no stones = 15 no. of fruits that have stones = 5 where monkey is picked randomly from the remaining 15. A). let the monkey eats five avocados, so if x is the number of avocados have stones. F(x) = k_ex(N - k)_c(n - k)/N_ek x = 0 f(x) = 0.1937 x = 1 f(x) = 0.4402 x = 2 f(x) = 0.2935 x =3 f(x) = 0.0677 x = 4 f(x) = 0.0048 x = 5 f(x) = 0.0001 let there are 150 avocado left. Let for any x , no. of avocado with stones left in remaining fifteen (5 - x) p_r(Avocado contains astone) = 5/15 times p_r(x = 0) + 4/15 p_r(x = 1) + 3/15 times p_r(x = 2) + 2/15 p_r(x = 3) + 1/15 p_r(x = 4) + 0/15 p_r(x = 5) = 5/15(o.1937) + 4/15(0.4402) + 3/15(0.2935) + 2/15(0.0677) + 1/15(0.0048) + 0/15(0.0001) =0.25 b) where p_r (Monkey monkey consumed at least one stone of Avocado contains a stone) = 1 - p_r(Monkey didn't consume stone 1 avacado contains a stone 1 - (5/15 * (0.1937))/0.25 = 0.7417