# Aptitude - Probability - Discussion

Discussion Forum : Probability - General Questions (Q.No. 2)
2.
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
 10 21
 11 21
 2 7
 5 7
Explanation:

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2 `
 = (7 x 6) (2 x 1)
= 21.

Let E = Event of drawing 2 balls, none of which is blue. n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls.
= 5C2
 = (5 x 4) (2 x 1)
= 10. P(E) = n(E) = 10 . n(S) 21

Discussion:
106 comments Page 2 of 11.

Krish said:   3 years ago
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.
(6)

ABI said:   3 years ago
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.

Summi said:   3 years ago
How come 7*6/2*1? please clear me.

Tharini said:   4 years ago
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.

HEMALATHA said:   4 years ago
@Pranav.

Magnus The Great said:   5 years ago
@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.

Christiano ronaldo said:   5 years ago
I didn't understand this. Please, anyone, help me to get it.

Arnika said:   5 years ago
What is C and S please explain?

Manish Kumar Gupta said:   5 years ago
How come this 2c2/7c2?

Harry said:   5 years ago
Can someone help me with this part of the question:

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7.
= 7C2 `
= (7 x 6).
(2 x 1),
= 21.

I don't understand what 7C2 means or how we get 7x6 or 2x1 then divided.
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