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?

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) =

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) =

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) =

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) =

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) =

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) =

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) =

understand that i don't know the answers

will someone answer his question