Aptitude - Probability - Discussion

Choose the first ball: There are 7 balls in total (2 red, 3 green, and 2 blue) , so you have 7 choices for the first ball.

Choose the second ball: After picking the first ball, there are 6 balls left for the second ball (since you're not concerned with color at this point).

Now, multiply the number of choices for the first and second balls together to get the total number of ways to draw any two balls: 7 choices × 6 choices = 42 ways.

Finally, calculate the probability that none of the balls drawn is blue by dividing the number of ways to draw without blue balls by the total number of ways to draw any two balls:

Probability = (Number of ways to draw without blue balls) / (Total number of ways to draw any two balls).
Probability = 20 / 42.

You can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:

Probability = (20 ÷ 2) / (42 ÷ 2).
Probability = 10 / 21.
So, the probability that none of the balls drawn is blue is 10/21.

See there are two ways to solve this question:

METHOD 1:

Probability of getting no blue balls in 2 draws = probability of getting 2 other balls in 2 draws out of any other 5 balls.

i.e., select two balls from the remaining 5 balls in 5C2 ways = n(E),

Now the sample space is n(S) = selection of 2 balls out 7b balls = 7C2,

So, required probability= n(E)/n(S),

=> 5C2/7C2 = [5!/(2!*3!)]/[7!/(2!*5!)]

=> 10/21 is the answer.

METHOD 2:

Probability of not getting any blue balls in two draws = 1 - (probability of getting 1 blue ball and 1 other ball + probability of getting 2 blue balls)

So,

Selection of 1 blue ball and one other ball = 2C1 * 5C1 = 10 ways,

Selection of 2 blue balls = 2C2 = 1,

So, probability = 1 - [{10/7C2} + {1/7C2}]

=> 10/21 answer.

@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.

If you are not aware of combinations like 7C2 then give names to the balls like r1, r2, g1, g2, g3, b1 and b2.

So if you take two balls out of the bag then if we consider r1 comes then total combination would be r1r2, r1g1, r1g2.. and so on. That means, for r1 there are 6 combinations.

So we can conclude following table.

Ball combination
r1 6
r2 5
g1 4
g2 3
g3 2
b1 1
b2 0

So total number of combination = 6+5+4+3+2+1 = 21.
And combination using blue ball = 2+2+2+2+2+1 = 11.

So combination without using blue ball = 21-11 = 10.

So probability of the same = 10/21 :).

When we say 6C subscript 2, it means
6 * 5 * 4 * 3 * 2 * 1 divided by 2 * 1.

So, the correct answer to me is 5/7.
Since, 2 red, 3 green, and 2 blue, are present,
Therefore,
S=7 (numbers of the balls)
Therefore,
Let E be the event space that blue ball is been chosen at random.
n(E)=2
n(s)=7 (total numbers of balls in the bag.
Pr(E)=n(E)/n(s)
=2/7.

To find the probability that none of the blue ball is drawn, 1 must be subtracted from the probability that the blue ball was chosen at random.

1 - 2/7.
7-2/7.
= 5/7.

Given criteria - Need the outcome to be either Red Or Green, but not Blue.

So, at first - The Total possible combination outcomes(Events) we would get considering Red Or Green(2+3 = 5) i.e "5c2".

Sample space(All Outcomes) = (Red+Green+Blue = 7) ; "7c2
Pr(E) = n(E)/n(S).

= 5c2 / 7c2,
=((5*4)/(2*1)) / ((7*6)/(2*1)),
=(20/2) /(42/2),
=(20/42),
=10/21,
=0.48 [ As probability defines, the chances of any Event occurrence is "Greater than or equal to '0' Or Less than or equal to '1'] [>=0 <=1].

Given criteria - Need the outcome to be either Red Or Green, but not Blue.

So, at first - The Total poswsible combination outcomes(Events) we would get considering Red Or Green(2+3 = 5) i.e "5c2"

Sample space(All Outcomes) = (Red+Green+Blue = 7) ; "7c2
Pr(E) = n(E)/n(S).
= 5c2 / 7c2,
=((5*4)/(2*1)) / ((7*6)/(2*1))
=(20/2) / (42/2)
=(20/42)
=10/21
=0.48 [ As probability defines, the chances of any Event occurrence is "Greater than or equal to '0' Or Less than or equal to '1'] [>=0 <=1].

Because, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.

Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
can u please explain how 2 comes here?.""""

But the question says NO BLUE BALL..why are we considering 1 blue and 1 green ...or 1 redor 1 blue..?

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.

The probability of the first ball not being blue is 5/7 and the probability of the second ball also not being blue is 4/6 and this is so because we have already drawn one ball and now the total number of balls got reduced to 6. Now where the 4 comes from? Its because we're assuming that the first ball we've drawn is a non-blue and therefore we have one less ball than 5 to draw from now.

I hope it was helpful.