Each time, you take out one ball from the box and then put it
back to the box. What is the probability that the number on the
ball you take out the first time is numbered less than 4, the
number on the second ball you take is numbered less than 2
and the number on the third ball you take is exactly 6? Enter
your answer in the form a/b.

40/3087

Step-by-step explanation:

First, we need to count the number of each type of numbered ball.

Ball Counts: 1 - 5 balls; 2 - 2 balls; 3 - 1 ball; 4 - 2 balls; 5 - 4 balls; 6 - 3 balls; 7 - 1 ball; 8 - 2 balls; 9 - 1 ball, there are then a total of 21 balls

Next, you need to find the probability of each event.

Get less than 4

p(<4) = (5+2+1)/21

p(<4) = 8/21

Get less than 2

p(1) = 5/21

Get exactly 6

p(6) = 3/21

p(6) = 1/7

Next, we need to find the probability of all three events by multiplying the probabilities together.

p(all three) = (8/21)*(5/21)*(1/7)

p(all three) = 40/3087

The probability of all three events occurring is 40/3087.

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer

did you get the answer yet?

step-by-step explanation: