The probbaility that the sample conatins at least one green ball is 0.90.
Let E be the event that denotes that the sample conatins at least one green.
According to the given question.
Total number of balls in an urn = 11 + 5 = 16
Total number of white balls = 11
Total number so green balls = 5
Now, the number of ways of selecting 7 balls from 16 balls
= 
= 16!/(7!9!)
= 11440
And the total number of ways of selecting 7 balls which contain atleast on green ball = 
= 
= 452 × 5 + 10 × 452 + 330 × 10 + 165×1 + 55 ×1
= 2260 + 4520 +3300 + 165 + 55
= 10300
Thereofore, the probability that the sample conatins at least one green ball is given by
P(E) = (total number of ways of selecting 7 balls which contain atleast on green ball) / (the number of ways of selecting 7 balls from 16 balls )
⇒ P(E) = 10300/11440
⇒ P(E) = 0.90
Hence, the probbaility that the sample conatins at least one green ball is 0.90.
Find out more information about probability here:
brainly.com/question/11234923
#SPJ4