Question

A set of pool balls contains 15 balls numbered 1-15.

Without replacement: What is the probability that an odd number ball is picked
out of a box twice without the first one being replaced?
With replacement: What is the probability that an even number ball is picked with
the first ball drawn being inserted back into the box?

Answer 1

Explanation:

a probability is always

desired cases / totally possible cases

the first case I assume means that we need the probability to pick 2 odd-numbered balls in a row, if we do not put the first drawn ball back into the box.

starting condition :

15 basks in total.

1, 3, 5, 7, 9, 11, 13, 15 = 8 odd numbered balls

2, 4, 6, 8, 10, 12, 14 = 7 even numbered balls

the probability for the first ball to be odd numbered :

8/15

now we have

14 remaining balls in total.

7 remaining odd numbered balls.

the probability of the second ball being odd numbered is

7/14 = 1/2

so, the probability of both as one combined event is

8/15 × 1/2 = 4/15 = 0.266666666...

now back to the starting condition.

the probability to pick an even numbered ball is

7/15

we put the ball back in and pull a second time.

the probability to an even numbered ball is

7/15

so, the probability of both as one combined event is

7/15 × 7/15 = 49/225 = 0.217777777...