Answer: A) 2/11
Step-by-step explanation:
P(green) = 5 green marbles out of 11 total marbles = [tex]\dfrac{5}{11}[/tex]
P(black) = 4 black marbles out of 10 remaining marbles = [tex]\dfrac{4}{10}[/tex]
P(green) and P(black) - "and" means to multiply them
[tex]\dfrac{5}{11}\times \dfrac{4}{10}=\dfrac{20}{110}\quad = \large\boxed{\dfrac{2}{11}}\quad when\ simplified[/tex]
The probability of selecting a green marble for the first trial and black marble for the second trial is 2/11. Hence, option A is the right choice.
The probability of any event A is the chance of the occurrence of that event.
It is determined as the ratio of the number of favorable outcomes to event A (n), to the total number of possible outcomes (S).
Hence, the formula for the probability of event A is given as:
P(A) = n/S.
In the question, we are informed that a student selects a marble from a bag, keeps it, and selects another. The bag contains 5 green marbles, 4 black marbles, and 2 blue marbles.
We are asked to find the probability of selecting a green marble for the first trial and black marble for the second trial.
Let the event of selecting a green marble on the first trial be A.
The number of outcomes favorable to the event A (n) = 5 {the number of green marbles}.
The total number of outcomes (S) = 11 {The total number of marbles in the bag}.
Therefore, the probability of event A is,
P(A) = n/S = 5/11.
Let the event of selecting a black marble on the second trial be B.
The number of outcomes favorable to the event B (n) = 4 {the number of black marbles}.
The total number of outcomes (S) = 10 {The total number of marbles in the bag. It is one less as one marble has been taken out}.
Therefore, the probability of event B is,
P(B) = n/S = 4/10.
Now, the probability of selecting a green marble for the first trial and black marble for the second trial is the product of P(A) and P(B).
Therefore, the probability = P(A) * P(B) = 5/11 * 4/10 = 2/11.
Learn more about the probability of an event at
brainly.com/question/7965468
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