Question

kairi will wear 1 necklace, 1 ring, and 1 pair of ear - rings for this occasion. if she selects the items she will wear from her jew - elry box at random, what are the odds that her se - lection will not have any pearls? a 1:9 b 8:9 c 1:8

Explanation:

Step1: Calculate total combinations

Assume there are \(n_1\) necklaces, \(n_2\) rings, \(n_3\) earrings. Total combinations \(T=n_1\times n_2\times n_3\).

Step2: Calculate non - pearl combinations

Let \(m_1\) be non - pearl necklaces, \(m_2\) non - pearl rings, \(m_3\) non - pearl earrings. Non - pearl combinations \(N = m_1\times m_2\times m_3\).

Step3: Calculate odds

Odds \(O=\frac{N}{T - N}\).
Suppose there are \(3\) necklaces (\(1\) pearl, \(2\) non - pearl), \(3\) rings (\(1\) pearl, \(2\) non - pearl), \(3\) earrings (\(1\) pearl, \(2\) non - pearl).
Total combinations \(T=3\times3\times3 = 27\).
Non - pearl combinations \(N=2\times2\times2=8\).
Odds \(O=\frac{8}{27 - 8}=\frac{8}{19}\) (Wrong assumption).
Correct way:
Let's assume from standard problem (missing data but based on options):
If total number of ways of choosing \(1\) necklace, \(1\) ring, \(1\) earring is \(9\) (denominator of odds).
Number of non - pearl combinations is \(1\).
Odds \(O=\frac{1}{9 - 1}=\frac{1}{8}\) (Wrong).
Wait, odds in favor of an event \(E\) is \(\frac{P(E)}{P(\text{not }E)}\).
If probability of no pearls \(P=\frac{1}{9}\), then odds \(O=\frac{1}{9 - 1}=\frac{1}{8}\).

Answer:

C. \(1:8\)