g(1 point) Two marbles are drawn randomly one after the other without replacement from a jar that contains 3 red marbles, 3 white marbles, and 6 yellow marbles. Find the probability of the following events, accurate to 4 decimal places. (a) A red marble is drawn first followed by a white marble. The probability is : (b) A white marble is drawn first followed by a white marble. The probability is : (c) A yellow marble is not drawn at all. The probability is :

Respuesta :

Answer:

(a) [tex]\frac{3}{44}[/tex]

(b) [tex]\frac{1}{22}[/tex]

(c) [tex]\frac{5}{22}[/tex]  

Step-by-step explanation:

We are given that Two marbles are drawn randomly one after the other without replacement from a jar that contains 3 red marbles, 3 white marbles, and 6 yellow marbles.

So, total number of marbles = 3 + 3 + 6 = 12 marbles

(a) Probability that a red marble is drawn first followed by a white marble is given by;

First P(drawing red marble) = [tex]\frac{3}{12}[/tex]

Now, since one red marble has been drawn so now the total marbles remaining will be 12 - 1 = 11 marbles and from this white marble will be drawn.

P(drawing white marble) = [tex]\frac{3}{11}[/tex]

Therefore, Probability that a red marble is drawn first followed by a white marble = [tex]\frac{3}{12} * \frac{3}{11}[/tex] = [tex]\frac{3}{44}[/tex] .

(b) Probability that a white marble is drawn first followed by a white marble is given by;

P(drawing first white marble) = [tex]\frac{3}{12}[/tex]

Now, the white marbles remaining are 3 - 1 = 2 marbles and total number of marbles are 12 - 1 = 11 marbles.

So, P(drawing another white marble) = [tex]\frac{2}{11}[/tex]

Therefore, Probability that a white marble is drawn first followed by a white marble = [tex]\frac{3}{12} * \frac{2}{11}[/tex] = [tex]\frac{1}{22}[/tex] .

(c) Probability that a yellow marble is not drawn at all is given by;

Now both the marbles will be drawn from red and white marbles i.e. 6   marbles and total marbles will be 12

So, probability that a yellow marble is not drawn at all = [tex]\frac{^{6}C_2 }{^{12}C_2}[/tex] = [tex]\frac{\frac{6!}{2!*4!} }{\frac{12!}{2!*10!} }[/tex]

                        = [tex]\frac{6!}{4!} * \frac{10!}{12!}[/tex] = [tex]\frac{6*5*4!}{4!} * \frac{10!}{12*11*10!}[/tex] = [tex]\frac{30}{132}[/tex] = [tex]\frac{5}{22}[/tex] .