Respuesta :
Answer:
The probability is : 0.90625
Step-by-step explanation:
4 balls are drawn with replacement.
There are total 4 balls so, total ways will be = [tex]4^{4}[/tex] = 256
Now, a green ball can be selected in 4C1 ways that is [tex]\frac{4!}{1!(4-1)!}[/tex] = 4 ways
Red ball can be selected in 3C1 ways = 3 ways
Yellow ball can be selected in 2C1 ways = 2 ways
And white ball can be selected in 1C1 way = 1 way
Now P(at least one color twice) = 1-P(no color repeat twice)
So, P(no color repeat twice) = [tex]\frac{4\times3\times2\times1}{4^{4} }[/tex]
= [tex]\frac{24}{256} =0.09375[/tex]
P(at least one color twice) = [tex]1-0.09375 =0.90625[/tex]
Hence, the probability is 0.90625.
The probability that there is at least one color that is repeated at least twice is 0.90.
Given
An urn contains 1 green ball, 1 red ball, 1 yellow, and 1 white ball.
I draw 4 balls with replacement.
How to determine the probability that there is at least one color that is repeated at least twice?
The probability that there is at least one color that is repeated at least twice is;
= 1 - the probability of no color repeats twice
The probability of selecting 1 green ball is;
[tex]\rm =\ ^4C_1\\\\= 4 \ ways[/tex]
The probability of selecting red ball is;
[tex]\rm = \ ^3C_1\\\\= 3 \ ways[/tex]
The probability of selecting a yellow ball is;
[tex]= \ \rm ^2C_1\\\\= 2 \ ways[/tex]
The probability of selecting a white ball is;
[tex]\rm = \ ^1C_1 \\\\ = 1 \ ways[/tex]
The total number of ways selecting 4 balls is;
[tex]= 4^4\\\\= 256[/tex]
Then,
The probability of no color repeats twice times;
[tex]= \dfrac{4 \times 3\times 2 \times 1}{256}\\\\= \dfrac{24}{256}\\\\=0.09[/tex]
Therefore,
The probability that there is at least one color that is repeated at least twice is;
= 1 - the probability of no color repeats twice
= 1-0.09
= 0.90
Hence, the probability that there is at least one color that is repeated at least twice is 0.90.
To know more about Probability click the link given below.
https://brainly.com/question/795909
