Question

A bag contains 9 red marbles, 5 white marbles, and 6 blue marbles. You draw 4 marbles out at random, without replacement. Find the following probabilities and round to 4 decimal places.

Answer 1

a) 0.0260

b) 0.0681

Explanation:

a)

The bag contains:

r = 9 (number of red marbles)

w = 5 (number of white marbles)

b = 6 (number of blue marbles)

So, the total number of marbles in the bag at the beginning is:

$n=9+5+6=20$

At the 1st attempt, the probability of choosing a red marble is:

$p(r)=\frac{r}{n}=\frac{9}{20}$ (1)

At the 2nd attempt, the 1st red marble is not placed back, so now the number of marbles is (n-1), while the number of red marbles left is (r-1). So the probability of choosing another red marble at the 2nd attempt is

$p(r)=\frac{r-1}{n-1}=\frac{8}{19}$ (2)

With a similar argument, the probabilities of selecting a red marble in the 3rd and 4th attempt are

$p(r)=\frac{7}{18}$ (3)

$p(r)=\frac{6}{17}$ (4)

Therefore, the probability of drawing 4 red marbles in the first 4 attempts without replacing is:

$p(rrrr)=\frac{9}{20}\frac{8}{19}\frac{7}{18}\frac{6}{17}=\frac{3024}{116280}=0.0260$

b)

At the 1st draw, the probability that the marble is not red is:

$p(r^c)=1-p(r)=1-\frac{9}{20}=\frac{11}{20}$

At the 2nd draw, there are 9 red marbles left and 19 total marble left. So, the probability of NOT drawing a red marble is:

$p(r^c)=1-\frac{9}{19}=\frac{10}{19}$

At the 3rd draw, there are 9 red marbles left and 18 total marbles left. So, the probability of NOT drawing a red marble is:

$p(r^c)=1-\frac{9}{18}=\frac{9}{18}$

Finally, with a similar argument the probability of NOT drawing a red marble at the 4th and last attempt is:

$p(r^c)=\frac{8}{17}$

So, the total probability of drawing 4 non-red marbles in the first 4 attempts is:

$p(r^cr^cr^cr^c)=\frac{11}{20}\frac{10}{19}\frac{9}{18}\frac{8}{17}=\frac{7920}{116,280}=0.0681$