Question 1174468: Whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, five green ones, two white ones, and two purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms.
She has no more than one white one.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $R, G, W, P$ be the number of red, green, white, and purple marbles in the bag, respectively.
We have $R=3$, $G=5$, $W=2$, $P=2$. The total number of marbles is $3+5+2+2=12$.
Suzan grabs 8 marbles.
The total number of ways to grab 8 marbles from 12 is $\binom{12}{8} = \binom{12}{4} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} = 495$.
We want to find the probability that she has no more than one white marble. This means she has either zero white marbles or one white marble.
Case 1: Zero white marbles.
If she has zero white marbles, she grabs 8 marbles from the remaining 10 marbles (3 red, 5 green, 2 purple).
The number of ways to do this is $\binom{10}{8} = \binom{10}{2} = \frac{10 \cdot 9}{2 \cdot 1} = 45$.
Case 2: One white marble.
If she has one white marble, she grabs 7 marbles from the remaining 10 marbles.
The number of ways to do this is $\binom{10}{7} = \binom{10}{3} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120$.
Since there are 2 white marbles, she can pick one of them in $\binom{2}{1} = 2$ ways.
So the number of ways to have one white marble is $120 \times 2 = 240$.
The total number of ways to have no more than one white marble is $45 + 240 = 285$.
The probability is $\frac{285}{495}$.
We can simplify this fraction.
Divide by 5: $\frac{285}{5} = 57$ and $\frac{495}{5} = 99$.
So we have $\frac{57}{99}$.
Divide by 3: $\frac{57}{3} = 19$ and $\frac{99}{3} = 33$.
So we have $\frac{19}{33}$.
Final Answer: The final answer is $\boxed{\frac{19}{33}}$
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