Question

From a box containing 4 black balls, 2 green balls, and 1 white ball, 3 balls is drawn in succession, each ball is not replaced in the box before the next draw is made. If X is the number of green balls drawn.

(a) What are the values for the set of x?
(b) What is the probability of having 2 green balls? (Fraction form)
(c) Make a probability distribution table.

Answer 1

Well let`s explain my solution:

a) It is clear that x has interval [0,2]. if you draw three balls, then there may be not be a green ball or may have maximum 2 balls.

b) In a box we have 7 balls and 2 of them are green. if we divide number green balls to total number of balls, we will find probablity of that at least one green ball was drawn: I meant this--> $\frac{2}{7}$ and for the second green ball: $\frac{1}{6}$ because first green ball took out. If we multiply both of them, we will find the probability of having 2 green balls: $\frac{2}{7}$ $\times$ $\frac{1}{6}$ = $\frac{1}{21}$ --> this is answer.

c) Let`s construct a probability distribution table.

x | white | green | black

P(X=x) | $\frac{1}{7}$ | $\frac{2}{7}$ | $\frac{4}{7}$ ---> this is distribution table.