Question

a bag contains eleven equally sized marbles, which are numbered. two marbles are chosen at random and replaced after each selection. what is the probability that the first marble chosen is shaded and the second marble chosen is labeled with an odd number? 10/121 24/121 6/11 10/11

Explanation:

Step1: Calculate the probability of choosing a shaded marble

There are 5 shaded marbles (\(1\), \(3\), \(4\), \(9\)) out of 11 marbles.
The probability \(P(\text{shaded})\) is \(\frac{5}{11}\).

Step2: Calculate the probability of choosing an odd - numbered marble

There are 6 odd - numbered marbles (\(1\), \(3\), \(5\), \(7\), \(9\), \(11\)) out of 11 marbles.
The probability \(P(\text{odd})\) is \(\frac{6}{11}\).

Step3: Use the multiplication rule for independent events

Since the marbles are replaced after each selection, the two events are independent.
The multiplication rule for independent events \(A\) and \(B\) is \(P(A\cap B)=P(A)\times P(B)\).
Let \(A\) be the event of choosing a shaded marble and \(B\) be the event of choosing an odd - numbered marble.
\(P(A\cap B)=\frac{5}{11}\times\frac{6}{11}=\frac{30}{121}\) (Wait, no. Wait, re - check the marbles:
Shaded marbles: \(1\), \(3\), \(4\), \(9\) (4 marbles). Odd - numbered marbles: \(1\), \(3\), \(5\), \(7\), \(9\), \(11\) (6 marbles)
\(P(\text{shaded})=\frac{4}{11}\), \(P(\text{odd})=\frac{6}{11}\)
\(P = \frac{4}{11}\times\frac{6}{11}=\frac{24}{121}\)

Answer:

\(\frac{24}{121}\) (the second option)