Question

A bag contains 3 white balls 4 black balls and 5 red balls if 3 balls are drawn at random determine the probability that.
1) all three red balls
2) all three are black
3) 2 are red and 1 is black
4) one of each Color is drawn
5) one is white
6) at least 1 white

Answer 1

Solution (1): The total number of outcomes when pulling a single ball (no matter what) is equal to the sum of all types of balls $N_1=3+4+5=12$. The total number of outcomes when pulling a red ball is the number of red balls $n_{r1}=5$ . The probability to get red ball first is $P_{r1}=n_{r1}/N_1=5/12$ . Before the second attempt of pulling out a red ball, the situation has changed according to

$N_2=3+4+4=11$ , $n_{r2}=4$ . Thus $P_{r2}=n_{r2}/N_2=4/11$ is a probability to get red ball in a second attempt. The probability of getting a red ball in the third attempt is also calculated $P_{r3}=3/10$. The total probability is a product of the probabilities of successful attempts:

$P_{(1)}=P_{r1}\cdot P_{r2}\cdot P_{r3}=\frac {5\cdot 4\cdot 3}{12\cdot 11 \cdot 10}=1/22$

Solution (2): Reasoning similarly to point (1) we get

$P_{(2)}=P_{b1}\cdot P_{b2}\cdot P_{b3}=\frac{4\cdot 3\cdot 2}{12\cdot 11\cdot 10}=1/55$

Solution (3): The total number of outcomes when pulling three balls out of 12 is equal to number of combinations $(^{12}_3)$which is convenient to write here as $C_{12}^3$ [1] $N=C^3_{12}=\frac{12!}{9!\cdot 3!}=220$

The total number of ways to pull two red balls out of 5 lying in the bag is equal to $n_{2r}=C_5^2$ . The total number of ways to pull one black ball out of 4 lying in the bag is equal to $n_{1b}=C_4^1=\frac {4!}{1!\cdot 3!}=4$ It was and just so clear. The probability of a positive result is estimated as a quotient of dividing the number of positive outcomes by the total number of cases

$P_{(3)}=\frac{n_{1b}\cdot n_{2r}}{N}=\frac{4\cdot C_5^2}{C_{12}^3}=\frac{4\cdot 5!/(2!\cdot 3!)}{12!/(3!\cdot 9!)}=\frac{4\cdot 5! \cdot 3!\cdot 9! }{12!\cdot 2!\cdot 3!}=2/11$

Solution (4): Using the results of the previous arguments we get $n_{1r}=C_5^1=5, n_{1b}=4, n_{1w}=3$ and $P_{(4)}=\frac{n_{1r}\cdot n_{1b}\cdot n_{1w}}{N}=\frac{5\cdot 4\cdot 3}{12!/(3!\cdot 9!)}=\frac{5\cdot 4\cdot 3\cdot 3!\cdot 9!}{12!}=$ 3/11

Solution (5): The total number of outcomes when pulling two balls not white is $n_{2rb}=C_{(12-3)}^2=C_9^2$ Thus $P_{(5)}=\frac{n_{1w}\cdot n_{2rb}}{N}=\frac{3\cdot C_9^2}{C_{12}^3}=\frac{3\cdot 9!/(2!\cdot 7!)}{12!/(3!\cdot 9!)}=\frac{3\cdot 9!\cdot 9!\cdot 3!}{2!\cdot 7! \cdot 12!}=27/55$

Solution (6): The number of outcomes when there are no white balls in the sample is equal to$n_{3rb}=C_{(12-3)}^3=C_9^3$ . To get the number of outcomes with any number of white balls, you need to subtract the obtained value $n_{3rb}$ from the total number of outcomes $N$ from previous. That is

$n_{w}=N-n_{3rb}$ . Finally we get

$P_{(6)}=\frac{n_w}{N}=1-\frac{n_{3rb}}{N}=1-\frac{C_9^3}{C_{12}^3}=1-\frac{9!/(3!\cdot 6!)}{12!/(3!\cdot 9!)}=1-\frac{9!\cdot 9!}{12!\cdot 6!}=1-21/55=34/55$

Answers:

1) $\frac{1}{22}$ ; 2) $\frac{1}{55}$ ; 3) $\frac{2}{11}$ ; 4) $\frac{3}{11}$ ; 5) $\frac{27}{55}$ ; 6) $\frac{34}{55}$

Note: $P_{(1)}=\frac{C_5^3}{C_{12}^3}; P_{(2)}=\frac{C_4^3}{C_{12}^3}$ ; $C_n^m=(^{n}_m)=\frac{n!}{m!\cdot (n-m)!}$

[1] https://en.wikipedia.org/wiki/Combination