has 10 red balls, 20 white balls. Probability takes 2 white balls.?

Question

a.4,4% b.16,7% c.44% d.54%

Answer 1

n(red)=10
n(white)=20
n(total)=30
event:2
probability will be =event happening in n(white)/event happening in n(total)
=20C2/30C2
=20*19/30*29
=38/87
=43.68 %
~44%

Answer 2

Probability of a white ball on the first draw = 20/30, which simply says you have 20 chances out of 30 possibilities of getting a white ball.

Probability of a white ball on the second draw is one of two possibilities:

1. If you drew a white on the first draw = 19/29
2. If you did not = 20/29

1. and 2. are conditional probabilities, where the condition is the outcome of the first draw. For both, the number of possibilities is 29, the thirty original balls minus the one you drew on the first draw.

But the number of white balls is different. If you drew one on the first draw, only 19 white balls remain. Otherwise, there are still 20 white balls out of the 29 that you might pick one out from.

The probability you draw two white balls, one each draw without replacement, on the first two draws is 20/30 X 19/29. This is your 44% answer.

Notice that all these probabilities are based on no replacement of a ball once it has been drawn. Notice also that you had to have drawn a white ball on the first draw. If you had drawn a red ball, the probability of drawing a white ball on two successive draws would have been zero. There is no chance of drawing two white ones in two draws when the first draw was red.

If there is replacement:

The probability you draw two white balls, one each draw with replacement, on the first two draws is 20/30 X 20/30. This is a bit over 45%.

Answer 3

Thinks is 44%

Answer 4

P of drawing 2 white balls from total 30 balls (with 20 white balls)

C(20,2)/C(30,2)
= 20.19/30.29
=38/87
=43.678% or 44%

Answer 5

44%

Answer 6

(20/30) = (2/3)

(2/3)(19/29) = (38/87)

ANS : .43678 or in your case c.) 44%

Answer 7

c - 44%

Answer 8

I don' know

Answer 9

43.56%?