in A, are you drawing only 3 balls?
then prob = (6/14)(5/13)4/12) = 5/91
or C(6,3)/C(14,3) = 20/364 = 5/91
in B, are you drawing only 2 balls?
then (8/14)(7/13) = 4/13
or C(8,2)/C(14,2) = 28/91 = 4/13
A golf bag contains 6 white balls and 8 yellow balls. What is the probability of each event?
A) Drawing 3 white balls?
B.) Drawing 1 Yellow ball and 1 white ball?
Thanks
3 answers
binomial distribution
p white = 6/14 = 3/7
p yellow = 1 - pwhite = 4/7
in three trials, what is the chance of drawing exactly 3 white?
The binomial coef of 3 out of three is 1
because 3!/(3!(3-3)!) = 1
P(3/3) = 1 * (3/7)^3 * (4/7)^0
so (3/7)^3
What is the probability of drawing exactly one white out of 2 trials?
C(2/1) = 2!/[1! (2-1)!] = 2
P(1/2) = 2 * (3/7)^1 * (4/7)^1
= 24/49
p white = 6/14 = 3/7
p yellow = 1 - pwhite = 4/7
in three trials, what is the chance of drawing exactly 3 white?
The binomial coef of 3 out of three is 1
because 3!/(3!(3-3)!) = 1
P(3/3) = 1 * (3/7)^3 * (4/7)^0
so (3/7)^3
What is the probability of drawing exactly one white out of 2 trials?
C(2/1) = 2!/[1! (2-1)!] = 2
P(1/2) = 2 * (3/7)^1 * (4/7)^1
= 24/49
Use what Reiny wrote. I ignored the changes in probability as the bag contained fewer balls.
4 answers