There are 98 balls in a box. 25 of them are red, 19 of them are green, 30 of them are purple, and 24 of them are blue. Suppose Hao draws 22 balls from the box with replacement (he draws the ball, records its color, and then puts it back into the box). Find the probability that he draws 2 red balls, 5 green balls, 10 purple balls, and 5 blue balls.

0,006

Step-by-step explanation:

The probability to choose 1 red ball is

The probability to choose 1 green ball is

19/98=0,19

The probability to choose 1 purple ball is

30/98=0,31

The probability to choose 1 blue ball is

24/98=0,24

We have replacement, so to choose for example 1 red ball at first is 0,25, and to choose again red ball is also 0,25. (Those are two independent events)

The probability to choose 2 red balls, 5 green balls, 10 purple balls and 5 blue balls (in this order) is:

0,25^2 x 0,19^5 x 0,31^10 x 0,24^5=0,065x0,0013x0,0000082x0,0008=55432e-13 (calculation)

But in question we don’t have order, we know just number of balls. So, we need to know how many orders we have with 2 red, 5 green, 10 purple and 5 blue balls. This is permutation with repeat. The number of orders is

(calculation)

So the answer is 55432e-13 x 1.0755e10=0,006 (claculation)

I’m not sure about calculation, but in the form of the answer i’m sure:)