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A juggler has a bag containing three yellow balls, six green balls, three blue balls, and one red ball, all the same size. The juggler pulls a ball from the bag at random. Then, without replacing it, he pulls out a second ball. What is the probability that the juggler first draws a green ball followed by a blue ball?

Respuesta :

lets put in the mind first we are finding a probability  for without replacement.

 Here we have 6 green,3 blue,1red balls
Probability of picking a green ball=Total Green/Total
which is going to be P1 =6/6+3+1=6/10=0.6

Now we are not putting back green ball in the bag again
 Probability of Blue ball=Total blue balls/Total(new)
                                     P2 =3/9=0.33

Now total will become 1 less than before as we did not put green ball back in the bag.

Now add them up =p1+p2=0.6+0.33=0.27

JeanaShupp

Answer:  [tex]\dfrac{3}{26}[/tex]

Step-by-step explanation:

Given: The number of green balls in bag =6

Total number of balls = [tex]3+6+3+1=13[/tex]

The probability of drawing first ball is green is given by :-

[tex]P(G)=\dfrac{6}{13}[/tex]

Since he didin't replace first ball, then the total number of remains = 12

The number of blue balls in the bag = 3

The probability of drawing second ball is blue is given by :-

[tex]P(B)=\dfrac{3}{12}[/tex]

Now, the probability that the juggler first draws a green ball followed by a blue ball is given by :-

[tex]P(G)\times P(B)=\dfrac{6}{13}\times\dfrac{3}{12}=\dfrac{3}{26}[/tex]

The probability that the juggler first draws a green ball followed by a blue ball : [tex]\dfrac{3}{26}[/tex]

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