a bag contains eleven equally sized marbles, which are numbered. two marbles are chosen at random and…

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

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

Answer

Explanation:

Step1: Calculate probability of first - event

There are 4 shaded marbles out of 11 marbles. The probability of choosing a shaded marble first, $P(A)$, is $\frac{4}{11}$ since probability $P=\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$.

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, $P(B)$, is $\frac{6}{11}$.

Step3: Calculate combined probability

Since the marbles are replaced after each selection, the two events are independent. For independent events $A$ and $B$, $P(A\cap B)=P(A)\times P(B)$. So $P(A\cap B)=\frac{4}{11}\times\frac{6}{11}=\frac{24}{121}$.

Answer:

$\frac{24}{121}$