A bag contains 3 black, 4 red, 3 yellow, and 2 green marbles. What is the probability of drawing a black and then a red marble out of the bag without replacing the black marble before drawing the red marble?

Respuesta :

ok so there is a total of 12 marbles. 

drawing a black is a probability of 3/12. drawing a red is a probability of 4:12

So first it is 3/12 and then because three are drawn instead of out of 12 it is out of 9 for red, so 4/9

3/12 x 4/9 

the answer is 1/9

You can use the fact that when you don't replace the marble, then the count of total marble is decreased.

The needed probability is [tex]\dfrac{7}{44}[/tex]

How to calculate probability for simple non weighted events?

To calculate the needed probability of an event for simple cases, we should first  count how many elements are in the favorable event. This is divided by total number of elementary events which gives the probability.

Elementary event is an event which contains only single outcome of the sample space.

How to calculate the needed probability in the given case?

Let the favorable event is denoted by E, then we have:

E = Drawing a black marble out of the bag and then without replacement, getting the red marble.

Since there are total 3+4+3+2 = 12 marbles, one marble can be taken out in [tex]^{12}C_1 = 12[/tex] ways.

After that there are 11 marbles remaining, out of which one marble can be taken out in [tex]^{11}C_1 = 11[/tex] ways.

Thus, by the chain rule of probability, we can do this in 12 times 11 = 132 ways to do so.

For favorable event, we need first marbleto be black marble and since there are 3 black marbles, this can be done in [tex]^3C_1 = 3[/tex] ways

Then there are two case. If first marble is black, then there are 4 red marbles remaining. From these marble marbles, we can take one red marble in [tex]^4C_1 = 4[/tex] ways.

If the first marble was red, then only 3 red marble marbles are remaining, thus, one red marble can be taken out in [tex]^3C_1 = 3[/tex] ways.

Thus total ways are 3 times (4+3) = 3 times 7 = 21 ways.

Thus, we have:

[tex]P(E) = \dfrac{\text{Number of elements in favorable events}}{\text{Number of elements in total}}\\\\P(E) = \dfrac{21}{132} = \dfrac{7}{44}[/tex]

Thus, the needed probability is [tex]\dfrac{7}{44}[/tex]

Learn more about probability here:

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