10 balls( 4red 3 blue 2 yellow 1 white). remove 2 balls. probability for 1 red and 1 yellow?

Question

a.8/45 b.8/35 c.2/3 d 10

Answer 1

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Answer 2

I'll assume that the ball removed first is not replaced before the second choice is made.

The answer is: 4/45.

Removing the red ball first, then the yellow.
chance for the red: 4/10 = 2/5
chance for the yellow: 2/9
chance for the red-yellow pair: (2/5)(2/9) = 4/45

Removing the yellow ball first, then the red.
chance for the yellow: 2/10 = 1/5
chance for the red: 4/9
chance for the red-yellow pair: (1/5)(4/9) = 4/45

Now I'll assume that the ball chosen first is replaced before the second choice is made.

The answer is: 2/25.

(4/10)(2/10) = 8/100 = 2/25.

Answer 3

2/5 X 2/9 = 4/45 X 2 = 8/45 (a.)

Answer 4

c 2/3

Answer 5

We have 2 cases Red, Yellow (RY) and Yellow, Red(YR). The required probability P = P(RY) + P(YR)
P(RY) = 4/10 * 2/9 = 8/90
P(YR) = 2/10 * 4/9 = 8/90
P 8/90 + 8/90 = 16/90 = 8/45
Therefore the answer is a.8/45

Answer 6

3/5

the prob of one of the balls being red + probability of one being yellow

4/10+ 2/10 = 6/10= 3/5

all assuming balls taken at the same time

Answer 7

red = 4/10
yellow = 2/9 (after one is removed)

4 x 2 = 8
10 x 9 = 90
8/90
answer is 4/45

Answer 8

(4/10) = (2/5)
(2/9)

(2/5)(2/9) = (4/45)

if it was Red then Yellow

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But if it was Yellow then Red

(2/10) = (1/5)
(4/9)

(1/5)(4/9) = (4/45)

So i don't see the actual answer up there.

Answer 9

(4/10)*(2/10)=8/100=2/25

Answer 10

You don't say if these are independant events. ie. are balls replaced after first choice, or 2 taken randomly?