Question

A bag contains four red marbles, three blue marbles, and three yellow marbles. You randomly pick three marbles without replacement. The first marble is red, the second marble is blue, and the third marble is red. Find the probability of this occurring.

Answer 1

4 red, 3 blue, 3 yellow ....total = 10

P(red) = 4/10 reduces to 2/5
without replacement
P(blue) = 3/9 reduces to 1/3
without replacement
P(red) = 3/8

P( all this happens) = 2/5 * 1/3 * 3/8 = 6 / 120 reduces to 1/20 <=

Answer 2

Answer: $\dfrac{1}{20}$

Step-by-step explanation:

Given : Number of red marbles in bag = 4

Number of blue marbles in bag = 3

Number of yellow marbles in bag = 3

Total marbles = $3+3+4=10$

Probability of the first marble is red :-

$\dfrac{4}{10}=\dfrac{2}{5}$

Probability of the second marble is blue :-

$\dfrac{3}{10-1}=\dfrac{3}{9}=\dfrac{1}{3}$

Probability of the third marble is red :-

$\dfrac{4-1}{9-1}=\dfrac{3}{8}$

Now, the probability of the first marble is red, the second marble is blue, and the third marble is red:-

$\dfrac{2}{5}\times\dfrac{1}{3}\times\dfrac{3}{8}=\dfrac{1}{20}$