Probabilities of draws without replacement

Question

A box contains 5 red balls and 7 orange balls. If Tim chooses 6 balls at random from the box, what is the probability that he will select 4 red balls and 2 orange balls?

Round your answer to 3 decimal places.

Answer 1

Two methods (using Excel's notation)

1) Combin(5,4)*Combin(7,2)/Combin(12,6) = 5 * 21 /924 = 5/44 = .113636

2) hypgeom.dist(4,6,5,12,0) = .113636 (in the population of size 12, we have 5 red; in the sample of 6, we have 4 reds)

Answer 2

There are a total of 12 balls (5 red and 7 orange balls) in the box. We will select 6 balls at random. We will use the probability with combinations.

1.) How many ways to select 6 balls at random? C(12, 6) = 12! / (6!)*(6!) = 924

2.) How many ways to select 4 red balls? C(5, 4) = 5! / (4!)*(1!) = 5

3.) How many ways to select 2 orange balls? C(7, 2) = 7! / (2!)*(5!) = 21

P(4 red and 2 orange balls) = (5*21) / 924 = 105/924 ≈ 0.114

The probability that Tim will select 4 red balls and 2 orange balls is about 0.114.

Answer 3

Since the process is without replacement and the order doesn't matter, the process involves combinations.

Different ways of drawing 6 balls out of the 12 is ¹²C₆ or 12!/(6!6!)=924 .

This represents the number in the sample space.

Different ways of chosing 4 red balls of 5 red balls and 2 orange balls of 7 orange balls is

⁵C₄ times ⁷C₂ = 5!/(4!1!) times 7!(2!5!)= 5 times 28 =140 This represents the number in the event space.

Probability of drawing 4 red balls and 2 orange balls = 140/924 which you can approximate to the appropriate number of decimal places.

Answer 4

This probability can be computed using the hypergeometric distribution principles.

First compute number of combinations of 5 red balls taken 4 at a time (5)

Then compute number of combination of 7 orange balls taken 2 at a time (21)

The number of combinations of both is 5 * 21 = 105

Then compute number of combinations of 12 balls taken 6 at a time (924)

Probability is 105/924 = 0.114