Aptitude - Probability - Discussion

@Pooja for A. White.
Ans = totql balls = 5+6+2+2 = 15.
Totql no. Of outcomes i.e. S=15c1 =15/1 = 15.
Event A = 5c1 = 5.
Probability of event A = 5/15 = 1/3.

What is the need for finding the first step? Please Explain.

Please solve this question:

A bag contains 5 white, 6 red, 2 green and 2 black balls. One ball is selected at random from the bag. Find the probability that the selected ball is;

A) white
B) non-white
C) white or green
D) black or red

Please solve this question:

A bank contains 5white,6red,2green, and 2black balls one ball is selected at random from the bag. Find the probability that the selected ball is:

1. White
2. Non white
3. White or green
4. Black or red

Not understanding, please Explain me this method.

How did you obtain 21 sample spaces in the above-mentioned solution?
Can you mention all possible combination of balls?

Ex:{ HH, TT, HT, TH } for 2coins.

Agree with you @Roma.

You are right Saurabh, in my way the method 1 is correct.

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.

Could anybody tell me what is that c means in 7c2?