Question

A bucket contains five green tennis balls, two yellow tennis balls, six red tennis balls, and eight blue tennis balls. Tony removes for tennis balls, without replacement from the bucket shown. What is the probability that Tony removes one yellow, one green ,one blue,and then one more blue tennis ball?

Answer 1

Answer:

The probability that Tony removes one yellow, one green ,one blue,and then one more blue tennis ball is 0.0039.

Step-by-step explanation:

We are given that a bucket contains five green tennis balls, two yellow tennis balls, six red tennis balls, and eight blue tennis balls.

As we know that, Probability of an event  =  $\frac{\text{Favorable number of outcomes}}{\text{Total number of outcomes}}$

Number of green tennis balls = 5

Number of yellow tennis balls = 2

Number of red tennis balls = 6

Number of blue tennis balls = 8

Total number of tennis balls = 5 + 2 + 6 + 8 = 21

{Now it should be notes that Tony removes for tennis balls, without replacement from the bucket means that every time the total number of remaining balls available for selecting will be one less}.

Now, Probability that Tony removes one yellow =  $\frac{2}{21}$

Probability that Tony removes one green =  $\frac{5}{20}$

Probability that Tony removes one blue =  $\frac{8}{19}$

Probability that Tony removes one blue again =  $\frac{7}{18}$

SO, final probability is =  $\frac{2}{21} \times \frac{5}{20} \times \frac{8}{19} \times \frac{7}{18}$  

                                      =  $\frac{2}{513}$

                                      =  0.0039