A) A bag contains 4 white balls, 3 red balls and 5 blue blues. A ball is selected and not put back. A second ball is selected. What is the probabilty that both balls are white?
B) Five Players line up, What is the probabailty that they are in aplha order by their names?
My answers:
A = 1/11
B = 1/120
2006-06-20 10:58:13 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
To get your answers, imagine what has to happen.
a) First, you have to pick 1 of the 4 white balls out of the 12 balls there.
2nd, you have to pick 1 of the 3 white balls left out of the 11 balls there.
So, in order to get a red and then a red, you need to get 4 out of 12 followed by 3 out of 11 or 4/12 times 3/11 or 12/132 = 1/22.
b) First, how many ways can 5 people line up? 5!. How many ways are correct (assuming they all have different names, only 1). So, the prob is 1/5! or 1/120.
2006-06-20 11:03:54 · answer #1 · answered by tbolling2 4 · 0⤊ 1⤋
A) The probably that the first ball was white is 4/12. Given that it was white, there are now 11 balls left, 3 of which are white. So the probability of getting a white on the second draw is 3/11. To find the probability of both, multiply 4/12 by 3/11, which gives 1/11. You are correct.
B) There are 5*4*3*2*1 combinations, or 120, only one of which would be alphabetical. So 1/120 is correct.
2006-06-20 18:02:44 · answer #2 · answered by -j. 7 · 0⤊ 0⤋
A=(4/12)*(3/11)=12/132=1/11
B=1/5!=1/120
2006-06-20 18:02:48 · answer #3 · answered by Cosmin C 2 · 0⤊ 0⤋
A.
Probability of getting a white ball the first time: 4/12 = 1/3
Prob. of getting a white ball the second time: 3/11
Prob. = 1/3 · 3/11
Prob. = 1/11
B.
prob. that the first one is in right place: 1/5
prob. that the 2nd one is in right place: 1/4
prob. that the 3rd one is in right place: 1/3
prob. that the 4th one is in right place: 1/2
prob. that the last one is in right place: 1/1
prob. that they are in proper order: (1/5)^5 = 1/120
^_^
2006-06-21 07:12:58 · answer #4 · answered by kevin! 5 · 0⤊ 0⤋
A.)
(4/12) = (1/3)
(3/11)
The prob that both are white is (1/3)(3/11) = (3/33) = (1/11) So you have this one correct.
B.)
nPr = (n!)/((n - k)!)
5P5 = (5!)/((5 - 5)!)
5P5 = 5!/1
5P5 = 5!
5P5 = 120
So yes you have this one correct
Both of your answers are correct.
2006-06-20 19:43:52 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
I can tell you A is wrong for sure.
They aren't asking what the odd are that the second ball picked would be white.
They are asking what are the odds that the two balls taken out would be white.
2006-06-20 18:07:06 · answer #6 · answered by jencanadian 3 · 0⤊ 0⤋
For A you are definitely correct.
If two people had the same name for B, you would have a problem, but other than that you are correct on both.
Good Job!
2006-06-20 19:06:44 · answer #7 · answered by quickster94 3 · 0⤊ 0⤋
A is right. The reasoning of "-j" is absolutly correct.
2006-06-20 18:22:49 · answer #8 · answered by johndersel 3 · 0⤊ 0⤋
Maybe it's me but a seems wrong. B is definantly right.
2006-06-20 18:10:59 · answer #9 · answered by Devilz Angel 3 · 0⤊ 0⤋