I'm having trouble finding how many ways there is to arrange 5 black, 7 red, 9 blue, and 6 white marbles to find the probability that every white marble is adjacent to at least one other white marble.
If you could help me out that would be great!
[Math] 5 black, 7 red, 9 blue, and 6 white marbles.
combinatoricsprobability
Related Solutions
There are a number of ways of approaching this problem, but the easiest solution is to realize that it doesn't matter what order you took the marbles out in. Take out a marble. Call it the "second" marble. Trivially then, the answer is $\frac{1}{3}$ since there is one white.
The other approach is more tedious. Consider what happens for the first marble. Either it is black or it is white.
The probability that it was black was $\frac{2}{3}$ and the probability that it was white was $\frac{1}{3}$
If the first was a black, there now remain one black and one white marble. The chance that the next marble drawn is white is now $\frac{1}{2}$. By multiplication principle then this occurs with probability $\frac{2}{3}\cdot \frac{1}{2} = \frac{1}{3}$
If the first marble was white, the second cannot be white (as there are no whites left in the urn/bag). Thus to arrive at this leaf of the probability tree by multiplication principle it occurs with probability $\frac{1}{3}\cdot 0 = 0$
The total probability then is the sum of these: $\frac{1}{3} + 0 = \frac{1}{3}$
In general, you will only add probabilities together if they are disjoint/mutually exclusive events.
$Pr(A\cup B) = Pr(A) + Pr(B) - Pr(A\cap B)$ and $Pr(A\cap B)=0$ whenever $A$ and $B$ are disjoint.
You will multiply probabilities for calculating intersections of events.
$Pr(A\cap B) = Pr(A)\cdot Pr(B|A)$ In the case that $A$ and $B$ are independent events, then $Pr(B|A)=Pr(B)$ in which case the intersection is simply the product $Pr(A)\cdot Pr(B)$.
The question asks for combinations, but permutation seems more intuitive to me in this question, as the order of the marbles being drawn has to be taken into account.
I would use P(60,2) as the sample space. By doing so, I have to treat all 60 marbles as distinctive:
Blue: Marble 1, Marble 2, Marble 3..., Marble 20
Green: Marble 21, Marble 22, Marble 23..., Marble 40
Red: Marble 41 to Marble 60
(a) drawing two marbles of the same colour
$Probability = \frac{20P2+20P2+20P2}{60P2}$
(b) drawing a blue and a green marble
$Probability = \frac{(20P1*20P1)+(20P1*20P1)}{60P2}$
(c) drawing a blue or a green marble
$Probability = \frac{(20P1*40P1)+(20P1*40P1)}{60P2}$
Look forward to an answer that uses combinations instead.
Best Answer
There are, in general, $n!$ ways to arrange $n$ objects. So you'd have $(5+7+9+6)!$ but then as all the black marbles are identical, their permutations shouldn't be counted. For every 'good' permutation, you also have another $5!7!9!6!$ that only differ from it by permutations of same colored marbles. Thus the result is $\frac{(5+7+9+6)!}{5!7!9!6!}$.
For the probability, consider that whites must be in pairs or triples. This is equivalent to solving the previous problem for $3$ and $2$ white marbles. Notice that the two cases overlap when all $6$ marbles are in one group.