10 balls( 4 red, 3 blue, 2 yellow s1 white). Remove 2 balls.probability for remove 2 red balls.?

Question

a.2/5 b.2/9 c.2/15 d.2/7

Answer 1

Probability for the first one being removed is 4 out of 10.

If the first one removed is red, then probability that the 2nd one is red is 3 out of 9. But if the first one removed is not red, then the probability that the 2nd one is red is 4 out of 9.

If the first and 2nd removed are not red, then the probability of the next one being red is 4 out of 8.

And so it goes!

Answer 2

If 2 balls are removed together, probability of removing red balls is 4/10 are red or that is 2/5 hence answer 'a' is correct

If first one has been picked out and is red, then probability of second one being red will be 3/9. Then the combined probability will be 2/5 multiplied by 3/9 = 6/45 or that is 2/15 hence answer 'c' will be correct.

Now the question is, are the two balls being removed simulataneously or not ??

Answer 3

Without replacing the ball before picking the second one
Probability of removing 1st red ball: 4/10
Probability of removing 2nd red ball: 3/9

So, 4/10 * 3/9 = 12/90 = 2/15 which is C

With replacing the ball before picking the second one
Probability will remain the same for both picks: 4/10

So, 4/10 * 4/10 = 16/100 = 4/25 which is not present in your answers so, it was asked for without replacement

Answer 4

2/5 x 1/3 = c 2/15

Answer 5

10 balls = 100%
4 balls out of 10 balls = 40 %
60 % of the balls are non red

answer A: 2/5 = 40 %

Answer 6

4/10 * 3/9 = 2/15. you have to take the probability of getting a red from the original set, them minus that ball from both the number of balls left, and the number of red balls left.

Answer 7

4/10 * 3/9 = 12/90 = 6/45 = c

Answer 8

If removed without replacement, prob 1st is 4/10, second is 3/9, compound probablility is the product, i.e. 2/15. So answer c is correct.

Answer 9

4/(4 + 3 + 2 + 1) = 4/10 = (2/5)
3/9 = (1/3)

(2/5)(1/3)

ANS : c. 2/15

Answer 10

Positively C