Question

A lab rat is given 6 randomly selected nuts from a bag of 20 nuts, of which 8 have poisonous chemicals.

a. What is the probability that none of the 6 nuts the rat eats are poisonous?

b. If the rat eats 3 poisonous nuts, it passes out. Let X be the number of poisoned nuts (where X cant be greater than 3). If f(x) is p.m.f of X, what is f(3)?

Answer 1

a) Probability that none of the 6 nuts that rat eats are poisonous is computed here as:

= Number of ways to select 6 nuts from the 12 non poisonous nuts / Total ways to select 6 nuts from 20 nutes

Therefore 0.0238 is the required probability here.

b) Given that the rat passes out with 3 poisonous nuts, the probability f(3) that is the rat eats poisonous nuts and passes out is computed here as:
= Probability that the first 3 nuts selected are poisonous + Probability that 2 nuts out of the first 3 nuts selected are poisonous * Probability that the fourth nut is poisonous + Probability that there are 2 poisonous nuts out of first 4 nuts * Probability that 5th nut is poisonous + Probability that there are 2 poisonous nuts out of first 5 nuts * Probability that 6th nut is poisonous

Therefore 0.4551 is the required probability here.