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There are 5 red marbles, 8 blue marbles, and 12 green marbles in a bag. What is the theoretical probability of randomly drawing a red marble and then a green marble? 10% 17% 9.6% 68%

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The answer is 10%!!!! I figured it out on my schoolwork cause I had the same exact question and I put what you said and I got it wrong...thx:( not.

the answer is actually 10%

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

[tex]P(E) = \dfrac{\text{Number of favorable outcome}}{\text{Number of total outcome}} = \dfrac{n(E)}{n(S)}[/tex]

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 [tex]^5C_1[/tex] ways,

and we can take out one marble out from 25 marbles in [tex]^{25}C_1[/tex] ways, thus,

[tex]P(\text{Taking one red ball out}) = \dfrac{^5C_1}{^{25}C_1}} = \dfrac{5}{25} = 0.2[/tex]

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 [tex]^{12}C_1[/tex] = 12 ways

and out of 24 marbles, we can take one marble in [tex]^{24}C_1 = 24[/tex] 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