Question

i need someone to help me with conditional probability... a total of 5 questions...

anyone up for the job?

Answer 1

it helps if you actually post the question .... that way people can know if they are capable of helping or not

Answer 2

well i jus need someone here that way i know someone will actually help haha but ill post the first one now

Answer 3

A bowl contains 25 balls numbered 1 to 25. A ball is drawn and its number is noted. Without replacing the first ball, another ball is drawn.
The probability that the numbers on both balls are odd numbers is............?

Answer 4

so how many balls are there to start with and how many of them are odd?

Answer 5

25 balls and 13 are odd so the probability of drawing an odd is 0.52... correct?

Answer 6

13/25 for the first draw yes

now the conditions have changed from the second draw; how many balls are left, and how many are odd given that an odd one has already been removed?

Answer 7

24 balls with 12 odd so the probability is 0.5..

Answer 8

good, so what is our probability of 2 odd balls in this case?

Answer 9

0.26 ?

Answer 10

13*12/(25*24)

13/50 ... yeah, .26

Answer 11

yyeeeaaa!!! haha so i got that one right... alright... shall we move on to the next one?

Answer 12

sure, but keep in mind that for study purposes its useful to post one question per thread.

Answer 13

alright well ill post a fresh one.. one minute..