Aptitude - Probability - Discussion

@ALL.

According to me,
2 red
3 green
2 blue
Total Balls =7 Balls.

Note: (Always solve as *without replacement* unless otherwise stated)

Pr(1st Red & 2nd red...RR) = 2/7 * 1/6 = 1/21,
Pr(1st Green & 2nd Green..RG)=3/7*2/6= 1/7,
Pr( 1st Blue & 2nd blue..BB) = 2/7 * 1/6 = 1/21,
Pr(1st Red & 2nd Green...RG) = 2/7 * 3/6 = 1/7,
Pr(1st Red & 2nd blue...RB) = 2/7 * 2/6 = 2/21,
Pr(1st Green & 2nd blue...GB) = 3/7 * 2/6 = 1/7,
Pr(1st green & 2nd red ...GR) = 3/7 * 2/6 = 1/7,
Pr(1st blue & 2nd red.... BR) = 2/7 * 1/6 = 1/21,
Pr(1st blue & 2nd green...BG) = 2/7 * 3/6 = 1/7.

THEN
Pr( None is Blue) = RR or GG or RG or GR.
1/21 + 1/7 + 1/7 + 1/7.
= 10/21.

@Pranav.

The answer is 62.

There is;

2 Red.
3 Green.
2 Blue Balls.
i.e. 5 Non-blue balls and 2 Blue.

1st ball can be picked from those 5 non-blue balls so that none of them is blue.

Hence n(E)=5 & n(S)= 7.
Hence probability is 5/7.

The next ball will be picked from remaining 6 balls and 4 Non-Blue balls(Since one is already picked).

Hence Probability(2nd picked ball not blue) = 4/6.

Hence,
Total probability = (5/7) * (4/6) = 10/21.

How come 7*6/2*1? please clear me.

The answer should be 5/7 right!

Because Probability = No.of favourable outcomes/Total outcomes.
=5/7.

Since, if both balls r not blue then they have to be either red(2) or green (3).
Thus,2+3 is 5.

So, option D is correct.

Using conceptual method rather than going by formula gives a better feel for what is happening.

P1 = probability of 1st ball not blue = 5/7 as 5 are non-blue.
P2 = probability of 2nd ball not blue = 4/6 as 1 non-blue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.

Simple explanation, well done, thanks @Krish.

(2R+3G)c2/7c2.
5c2/7c2.
5 * 2/2/7 * 6/2.
(20/2)/(42/2).
10/21 is the final answer

Total number of balls gives 2+3+2=7 s be the sample space.
number of drawing balls n(s)=2 balls out of 7.
7 c 2 = 7 * 6 = 42/2 = 21.

e = drawing balls none of which is blue,
2 balls out of 2+3=5.
5 c2.
5*4 = 20/2 = 10.

Therefore p n(e)= 10/21.

Can anyone tell me if I subtract the probability of getting 2 blue balls i.e 2/7 from 1 then why I can't get the answer?

Please explain me.