Probability

13. A bag contains 9 red marbles, 10 white, 5 blue. If 6 are drawn at random, what is the
probability
a) 3 blue
3 blue
3/6 or 1/2

b) same number of each colour

=9c2 x 10c2 x 5c2/24c6
=16200/134,596
=0.12



c) NONE WILL BE RED
9/24 is the probability of having a red.
Thus, 15/24 is the probability of not having a red



d) all 6 will be the same colour
=10c1x 9c1 x 8c1x 7c1 x 6c1 x5c1/24c1x23c1x22c1x21c1x20c1x19c1
=2.54x10^10
I am not sure if my answers are correct. Thanks for help!!

kay_wink08 said:

13. A bag contains 9 red marbles, 10 white, 5 blue. If 6 are drawn at random, what is the
probability
d) all 6 will be the same colour
=10c1x 9c1 x 8c1x 7c1 x 6c1 x5c1/24c1x23c1x22c1x21c1x20c1x19c1
=2.54x10^10

Click to expand...

Your answers have several problems.
The answer to part d) is:
\(\displaystyle \dfrac{\binom{9}{6}}{\binom{24}{6}}+\dfrac{\binom{10}{6}}{\binom{24}{6}}\)

Hello, kay_wink08!

13. A bag contains 9 red marbles, 10 white, 5 blue.
If 6 are drawn at random, what is the probability of:

a) (exactly) 3 blue

Click to expand...

There are: 5 Blue and 19 Others.
We want 3 Blue and 3 Others.

\(\displaystyle P(\text{exactly 3 Blue}) \:=\:\dfrac{(_5C_3)(_{19}C_3)} {_{24}C_6} \;=\;\dfrac{9,\!690}{134,\!596} \;\approx\;0.072\)

b) same number of each colour

\(\displaystyle \dfrac{(_9C_2)(_{10}C_2)(_5C_2)}{_{24}C_6} \;=\;\dfrac{16,\!200}{134,\!596} \;\approx\;0.12\)

Click to expand...

Correct!

c) none will be red.

Click to expand...

There are 9 Red, and 15 Others.
We want 6 Others.

\(\displaystyle P(\text{none of the 6 are red}) \:=\:\dfrac{_{15}C_6}{_{24}C_6} \;=\;\dfrac{5,\!005}{134,\!596} \;\approx\;0.037\)

d) all 6 will be the same colour

Click to expand...

\(\displaystyle \begin{array}{ccc}\text{6 Red:} & _9C_6 \:=\:84\\ \text{6 White:} & _{10}C_6 \:=\:210 \\ \text{6 Blue:} & \text{impossible} \end{array}\)

\(\displaystyle P(\text{6 same color}) \:=\:\dfrac{294}{134,\!596} \;\approx\;0.0022\)