# What is the probability that the person picks up all three red balls?

Let's say there are $10$ orange balls and $3$ red balls in a bag.

And a person picks a ball at random and puts it on the side. So for example if he picked up an orange ball then there would be $9$ orange balls and $3$ red balls left.

What is the probability that the person picks up all three red balls in three tries?

Would be $\frac1{14}\times\frac1{13}\times\frac1{12}$

• Are there three or four red balls in the bag – Plus Twenty Jun 16 '18 at 1:03
• There are three red balls – Bas bas Jun 16 '18 at 1:04
• If they picks up 3 balls and they all happen to be red – Bas bas Jun 16 '18 at 1:05

The probability of getting all red balls would be, if he is only grabbing balls three times:

$\frac3{13}\times\frac2{12}\times\frac1{11} = \frac1{286}$

Since the balls are being removed you must subtract 1 from both the numerator and denominator as the number of red balls and total amount of balls are decreasing every time a ball is removed.

This probability is $$\frac{\text{# ways of choosing 3 red balls (order matters)}}{\text{# ways of choosing 3 balls (order matters)}} = \frac{3 \times 2 \times 1}{13 \times 12 \times 11}.$$

A proper way to denote exactly what is described in the accepted answer of Conal.

For $i=1,2,3$ let $E_i$ denote the event that the $i$-th ball picked is red.

Then you are looking for:$$P(E_1\cap E_2\cap E_3)=P(E_1)P(E_2\mid E_1)P(E_3\mid E_1\cap E_2)=\frac3{13}\frac2{12}\frac1{11}$$

Another way to solve this is to realize that the $13\choose3$ subsets of $3$ balls of the urn’s $13$ balls are equally likely outcomes of drawing three balls without replacement and these subsets comprise all possible outcomes. Only $1$ of these subsets consists of three red balls, so the probability of obtaining this subset is $1\big/{13\choose3}$.

• Mmmm, are you sure there are 11 balls? – Tony Hellmuth Jun 16 '18 at 2:48
• Oops, and thanks! I fixed it. – Steve Kass Jun 16 '18 at 18:59