Question

One way to denote the number of combinations is $C_{n,k}$, but we shall instead use notation that is quite common in probability books: $C(n,k)$ or $\binom{n}{k}$, read "n choose k". $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ Problem 1. Suzy is given a bag containing 4 red marbles, 3 green ones, 2 white ones, and 1 purple one. She grabs five of them. Find the probabilities of the following events, expressing each as a fraction in lowest terms. $4R + 3G + 2W + 1P = 10$ marbles in the bag $C(10, 5) = 252 = n(S)$ a) She has all the red ones. $n(E) = C(4,4)C(6,1) = 6$ (4 red, 1 non-red) $P(E) = \frac{n(E)}{n(S)} = \frac{6}{252} = \frac{1}{42}$ b) She has at least one white one.

          One way to denote the number of combinations is $C_{n,k}$, but we shall instead use notation that is quite common in
probability books: $C(n,k)$ or $\binom{n}{k}$, read "n choose k". $\binom{n}{k} = \frac{n!}{k!(n-k)!}$
Problem 1. Suzy is given a bag containing 4 red marbles, 3 green ones, 2 white ones, and 1 purple one. She grabs five of
them. Find the probabilities of the following events, expressing each as a fraction in lowest terms.
$4R + 3G + 2W + 1P = 10$ marbles in the bag $C(10, 5) = 252 = n(S)$
a) She has all the red ones.
$n(E) = C(4,4)C(6,1) = 6$ (4 red, 1 non-red)
$P(E) = \frac{n(E)}{n(S)} = \frac{6}{252} = \frac{1}{42}$
b) She has at least one white one.

$One way to denote the number of combinations is Cn,k, but we shall instead use notation that is quite common in probability books: C(n,k) or nk, read "n choose k". nk = (n!)/(k!(n-k)!) Problem 1. Suzy is given a bag containing 4 red marbles, 3 green ones, 2 white ones, and 1 purple one. She grabs five of them. Find the probabilities of the following events, expressing each as a fraction in lowest terms. 4R + 3G + 2W + 1P = 10 marbles in the bag C(10, 5) = 252 = n(S) a) She has all the red ones. n(E) = C(4,4)C(6,1) = 6 (4 red, 1 non-red) P(E) = (n(E))/(n(S)) = (6)/(252) = (1)/(42) b) She has at least one white one.$

Added by Jacob J.

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