Question

There are 4 red balls and 6 green balls in a bag.
You reach in the bag and take out 3 balls without looking.
What is the probability that all three of the balls you take out are red?

Answer 1

Answer:

$\frac{1}{30}$

Step-by-step explanation:

From the question, in the bag there are;

4 red balls

6 green balls

10 balls in total.

Now, reaching in the bag and taking out 3 balls without looking, the probability that all three balls are red, can be analyzed as follows;

All three red means;

The first ball is red,

The second ball is red and;

The third ball is red.

i. First you take out a ball from a total of 10 balls. The probability P⁰(R) of having a red ball is given as;

P⁰(R) = $\frac{possible-space}{total-space}$

Since there are 4 red balls, the possible-space is 4

Also, since there are a total of 10 balls, the total-space is 10

P⁰(R) = $\frac{4}{10} = \frac{2}{5}$

ii. Secondly, you take out a ball from a remaining total of 9 balls. The probability P¹(R) of still having a red ball is given as;

P¹(R) = $\frac{possible-space}{total-space}$

Since there are 3 red balls remaining, the possible-space is 3

Also, since there are a remaining total of 9 balls, the total-space is 9

P¹(R) = $\frac{3}{9} = \frac{1}{3}$

iii. Thirdly, you take out a ball from a remaining total of 8 balls. The probability P²(R) of still having a red ball is given as;

P²(R) = $\frac{possible-space}{total-space}$

Since there are 2 red balls remaining, the possible-space is 2

Also, since there are a remaining total of 8 balls, the total-space is 8

P²(R) = $\frac{2}{8} = \frac{1}{4}$

Therefore, the probability P(R) of taking out three red balls without looking is given by the product of the probabilities described above. i.e

P(R) = P⁰(R) x P¹(R) x P²(R)

P(R) = $\frac{2}{5} * \frac{1}{3} * \frac{1}{4} = \frac{1}{30}$