Probabilities of draw without replacement

Question

An urn contains 3 white balls and 9 orange balls. If Sam chooses 8 balls at random from the urn, what is the probability that he will select white2 balls and 6 orange balls? Round your answer to 3 decimal places.

Answer 1

This is a hypergeometric distribution problem. There are a total of ₁₂C₈ ways to select 8 balls from the 12 balls. For the particular selection, there are a total of (₃C₂)(₉C₆) ways of doing it, so the probability is (₃C₂)(₉C₆)/₁₂C₈ = 0.5091.

Answer 2

Total number of Balls in Urn = 12 balls ( 3 white + 9 orange )

Probability Definition = Favorable number of cases / Total Number of cases

P = m / n ( Say )

Now n = Out of 12 balls 8 balls can be selected in ( 12 C 8 ) ways

12 C 8 = 12 !/ 8! ( 12-8)!

= 12 x 11 x 10 x 9 x 8 ! / 8! (8!)

= 11880/ 40320

= 0.294

Hence n = 0.294

Now m = out of 3 white balls 2white balls can be selected in (3C2) ways and out of 9 orange balls 6 orange balls can be selected in (9C6) ways

m = 3C2 X 9C6

Use the same formula we get 3 x 168 = 504

Hence m= 504

Therefore, the required probability is

P= m/n

= 504/ 0.294

P = 1714.285

Answer 3

The number of possible "draws" of 2 white balls from 3 white balls is given by ₃C₂ which equals 3!/2!/(3-2)! which translates to (3x2x1)/(2x1)/(1) or 3.

The number of possible "draws" of 6 orange balls from 9 orange balls is given by ₉C₆ which equals 9!/6!/(9-6)! which translates to (9x8x7)/(3x2x1) or 84.

The number of possible "draws" of 8 balls from 12 balls is given by ₁₂C₈ which equals 12!/8!/(12-8)! which translates to (12x11x10x9)/(4x3x2x1) or 495.

The probability of drawing exactly 2 white balls and 6 orange balls from 3 white balls and 9 orange balls is then equal to ₃C₂ x ₉C₆ ÷ ₁₂C₈ or

(3x84)/495 or 28/55 equal to 0.5090909091 and equivalent to 0.510.

Answer 4

There are a few key things to notice in this problem, the most important being that this is selection without replacement. This automatically disqualifies it from being a binomial distribution because, with binomial, each selection is independent with equal probability of success.

This requires using your combinatorial knowledge in conjunction with basic probability theory. It all boils down to how many ways each thing can happen.

We'll deal with the denominator first. How many ways can 8 balls be selected from the total 12? This is a combination, which is ₁₂C₈, or 495. This is the total number of possible outcomes and goes in the denominator.

So how do we count the successes? This is two distinct combinations occurring together.

The 2 white balls must come from the total 3 white balls. There are ₃C₂ ways that can happen, or 3.

The 6 orange balls must come from the total 9 orange balls. There are ₉C₆ ways that can happen, or 84.

Combine these 2 events, and there are 84 • 3 ways this composite event can occur, or 252. This is the total number of successful outcomes that are possible.

Put it all together, and the probability is ²⁵²/₄₉₅, or about 0.51.