You can use the fact that we're drawing the balls thus the numbers of balls will decrease.
The probability of randomly drawing a red marble and then a green marble is given by
Option A: 10%
How to find the probability for some event?
For simple cases, when events are not biased, we can use
\(P(E) = \dfrac{\text{Number of favorable outcome}}{\text{Number of total outcome}} = \dfrac{n(E)}{n(S)}\)
where S denotes sample space and the function n counts the number of items in the input set.
Using above definition
There are total of 5+8+12 = 25 marbles.
There are total of 5 red marbles,
Let the event E denotes event of randomly drawing a red marble and then a green marble.
Thus, we have
Since we can take out 1 red marble out of 5 red marbles in \(^5C_1\) ways,
and we can take out one marble out from 25 marbles in \(^{25}C_1\) ways, thus,
\(P(\text{Taking one red ball out}) = \dfrac{^5C_1}{^{25}C_1}} = \dfrac{5}{25} = 0.2\)
Since we're calculating the number of ways for favorable case the second marble can be taken out in:
Then there are 12 green marbles and thus, one green can be taken out in \(^{12}C_1\) = 12 ways
and out of 24 marbles, we can take one marble in \(^{24}C_1 = 24\) ways
Thus, P(taking green marble out) = 12/24 = 0.5
Thus, by chain rule of probability, we have:
probability of this case = P(E) = 0.2 times 0.5 = 0.1
Thus, we have P(E) = 0.1 = 10%
Thus,
The probability of randomly drawing a red marble and then a green marble is given by
Option A: 10%
Learn more about probability here:
https://brainly.com/question/1210781
