Help with probability.

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Json1289

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Can someone please let me know if I'm on the right track with this question -

A bag contains 4 white, 6 yellow, 5 blue, and 7 red balls. If a set of 4 balls is randomly selected, what is the probability that each of the balls will be
(a) of the same color if drawn with no replacement.
(b) of different color if drawn with no replacement.


For (a), I'm doing -
P(wwww) = 4/22 * 3/21 * 2/20 * 1/19
P(yyyy) = 6/22 * 5/21 * 4/20 * 3/19
P(bbbb) = 5/22* 4/21* 3/20* 2/19
P(rrrr) = 6/22 * 5/21 * 4/20 * 3/19

Then adding up the 4 probabilities above

For (b), I'm doing -
P(w) = 4/22
P(y) = 6/21
P(b) = 5/20
P(r) = 7/19

Then also adding up the 4 probabilities
 
(a) is good. You add because you want P(wwww or yyyy or bbbb or rrrr), and those are mutually exclusive.

For (b), what do you mean by P(w)? Is it that the first ball drawn is white? Couldn't it be second instead? And why add? This isn't an "or" situation.
 
i am confused that you have answered [MATH](a)[/MATH] correctly and [MATH](b)[/MATH] not correctly while [MATH](a)[/MATH] is more difficult than [MATH](b)[/MATH]
[MATH]\frac{4}{22} + \frac{6}{21} + \frac{5}{20} + \frac{7}{19} > 1 [/MATH] (what?????)

can you see now why it is wrong?
 
Yes I can see why it's wrong. Is it right to say that the answer to (b) is 1-(a)?
 
Looking at joshuaa post slightly differently.

All the denominators are about 20. Although we will be off a little lets assume they are all 20. Now when you add the numerators you get 22. Sure 22/20 > 1 but maybe our error caused it to be over 1. But in the end if you did not make a mistake it will sure be near 1 (and under). Do you really believe that p(of different color if drawn with no replacement) is really near 1?????
 
Json1289 said:
Can someone please let me know if I'm on the right track with this question -

A bag contains 4 white, 6 yellow, 5 blue, and 7 red balls. If a set of 4 balls is randomly selected, what is the probability that each of the balls will be
(a) of the same color if drawn with no replacement.
(b) of different color if drawn with no replacement.


For (a), I'm doing -
P(wwww) = 4/22 * 3/21 * 2/20 * 1/19
OR P(yyyy) = 6/22 * 5/21 * 4/20 * 3/19
OR P(bbbb) = 5/22* 4/21* 3/20* 2/19
OR P(rrrr) = 6/22 * 5/21 * 4/20 * 3/19

Then adding up the 4 probabilities above

For (b), I'm doing -
P(w) = 4/22
AND P(y) = 6/21
AND P(b) = 5/20
AND P(r) = 7/19

Then also adding up the 4 probabilities
Click to expand...
You need to learn to put AND or OR in your statements above. I did that for you. For AND you multiply, NOT add and for OR you add.
As Dr Peterson pointed out what you would have computed if you had multiplied the numbers in part b could have been P(1st = w and the 2nd = y and the 3rd = b and the 4th = r). But is that the only order?
 
Json1289 said:
Yes I can see why it's wrong. Is it right to say that the answer to (b) is 1-(a)?
Click to expand...
No, 1-(a) would be the probability that they are not all of the same color, not that they are all different. The former includes, say, wwwy, but the latter doesn't.

How many different orders for the four colors are possible if they are all different?
 
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