suppose there are 4 white balls
6 black balls
6red balls in a urn.
if 2 balls are selected from the urn without replacement,
what is the probability that the balls drawn are of the same color?
2006-08-28 15:09:02 · 5 answers · asked by tenstrike 2 in Science & Mathematics ➔ Mathematics
Think about this problem this way:
The probability you are looking for is the union of three events:
P(two whites) + P(two blacks) + P(two reds)
It is important to recognize that the balls are selected without replacement.
Solving for whites: 4/16 * 3/15 = 0.05
Solving for Blacks; 6/16* 5/15 = 0.125
Solving for Reds: 6/16*5/15 = 0.125
Solution is the Union of the three events or 0.30
Good Luck.
2006-08-28 15:28:57 · answer #1 · answered by alrivera_1 4 · 2⤊ 0⤋
Probability of
First Ball white = 4/16
And second ball white = 3 / 15
So both white = 4/16 * 3/15 = 12 / 240
Similarly for black balls 30 / 240 and
for red balls 30 / 240
So probability of ANY of the above = 12/240 + 30 / 240 + 30 /240
= 72 / 240 = 3 / 10 = 30 /100
Therefore there is a 30% chance.
2006-08-29 02:41:18 · answer #2 · answered by blind_chameleon 5 · 0⤊ 0⤋
You have 16 balls total, so the first one has 16 choices, and the second 15, for a total of 240.
Of those, there are 12 ways to get 2 whites, 30 ways to get two blacks and 30 ways to have 2 reds; so 72 ways to get two of the same color.
72/240 is 0.3, or 30% chances.
2006-08-28 22:15:55 · answer #3 · answered by Vincent G 7 · 1⤊ 0⤋
Fifty Fifty probability.
2006-08-28 22:15:36 · answer #4 · answered by Mr. BIG 6 · 0⤊ 0⤋
i think its 3 out of 4 times...i think....thats hard..
2006-08-28 22:14:38 · answer #5 · answered by hello taco. 3 · 0⤊ 0⤋