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? $\frac{10}{121}$ $\frac{24}{121}$ $\frac{6}{11}$ $\frac{10}{11}$

Explanation:

Step1: Calculate probability of first event

There are 5 shaded marbles out of 11 marbles. The probability of choosing a shaded marble first is $P_1=\frac{5}{11}$.

Step2: Calculate probability of second event

There are 6 odd - numbered marbles out of 11 marbles. The probability of choosing an odd - numbered marble second is $P_2=\frac{6}{11}$.

Step3: Use multiplication rule for independent events

Since the marbles are replaced after each selection, the two events are independent. The probability of both events occurring is $P = P_1\times P_2=\frac{5}{11}\times\frac{6}{11}=\frac{30}{121}$. But there is a mistake above, let's correct it. There are 4 shaded marbles out of 11 marbles, so $P_1=\frac{4}{11}$, and 6 odd - numbered marbles out of 11 marbles, $P_2 = \frac{6}{11}$. Then $P=\frac{4}{11}\times\frac{6}{11}=\frac{24}{121}$.

Answer:

$\frac{24}{121}$