# Aptitude - Probability - Discussion

### Discussion :: 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?

[A].
 10 21
[B].
 11 21
[C].
 2 7
[D].
 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

 Aarthi said: (Aug 16, 2010) Can we do this n different method? finding the possibility that both are blue and finally ans-1.

 Soundar said: (Aug 24, 2010) Its a correct answer and short way.

 Kk585 said: (Dec 24, 2010) Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21 Probability that none of ball is blue ---> 1-(1/21) = 20/21 Why is my answer wrong please explain? p(a)+p(a(bar)) = 1 .......

 Saurabh said: (Mar 22, 2011) This is the the simplest method to find the Probablity...........

 Rose Were said: (May 19, 2011) Why are you multiplying by 6 and by 4?

 Rose Were said: (May 19, 2011) How did the multiplications come about? please answer me.

 Subhash said: (Jun 3, 2011) What Kk585 said is not correct. Becuase, 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. Therefore total is 11/21. Hence ans is 1-(11/21) is 10/21.

 Ekta said: (Aug 8, 2011) Please clarify it which one having best answer of the solution.

 Akash said: (Sep 26, 2011) Can we do it by this way? 3c2+2c2+3c1 2c1 --------------- 7c2. I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green. Isn't the solution correct?

 Shiba said: (Sep 27, 2011) n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls. ? My question is why are you finding it only from 5 balls why not 7 balls.

 Srija said: (Nov 5, 2011) You are right akash!

 Dileep said: (Feb 14, 2012) The way akash did is exact. It takes into account all the probabilities

 Jay said: (Mar 2, 2012) (7 x 6) (2 x 1) Please explan it.

 Neha said: (Mar 29, 2012) What is 3c2, 2c2 etc? could anyone please explain this whole sum according to a class 7 student.

 Pranav said: (May 10, 2012) A bag contains 3 red, 2 green and 5 blue balls. 4 balls are drawn at random. What is the probability that 1 green ball?

 Madhu said: (Jul 8, 2012) Becuase, 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?

 Nishant said: (Aug 13, 2012) Can you please explain me why we select Number of ways of drawing 2 balls out of (2 + 3) balls instead of there having total no.of 7 balls, . So prob. Of drawing 2 balls must be out of 7 i.e. (2+3+2).

 Krish said: (Aug 17, 2012) 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..?

 Krish said: (Aug 17, 2012) I am not geting wats wrong in this...please explain.... Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21 Probability that none of ball is blue ---> 1-(1/21) = 20/21 Why is my answer wrong please explain? p(a)+p(a(bar)) = 1 .......

 Jaani! said: (Aug 23, 2012) @krish, Cuz 2C2 means picking those 2 blue balls out of those 2 blue balls only! Bt d ques says "None of them blue" bt one blue and one of any other colour cant b included also! Thats y we do that of picking up only green and red.. (no possiblty of blue)!

 Rajee said: (Sep 10, 2012) Here. No of out comes is 7c2 and no of events is 3c2+2c2+2c1. 3c1 then no of events possible is. P (E) =n (E) /n (s) =11/21. But we have to find here is none of the two ball is blue so we have to find. P (~E) =1-P (E) =10/21.

 Piyul said: (Nov 22, 2012) Is there any other way to understand basic concept of probability?

 Farah said: (Dec 5, 2012) Formula is probability is p(A) = n(A)/n(s), so we find n(s) = 7C2 = 21. Next we find n(a) = 5C2 = 10 these amounts put in formula p(A)=n(A)/n(S) Ans is 10/21 We have interested in 2 balls none in white so 2+5 not 2 white balls I think you understood

 Prabhat said: (Dec 15, 2012) 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.

 Kavita Kant said: (Mar 11, 2013) Is there any way to determine that when we take combination to find probability and when we take direct favourable events/total no of events. Actually I am confused in determining where to take combination and where not.

 Sagar said: (Mar 19, 2013) 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. A: Event none of the balls draw is blue. n(A): number of possible ways = 2C1*3C1+2C2*3C0+3C2*2C0 =2*3+1*1+3*1 =6+1+3 =10. and P(A)=n(A)/n(S)=10/21=1/2.

 Sandeep said: (Jul 31, 2013) 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 :).

 Navcool said: (Sep 4, 2013) Is it with or without replacement?

 Manjunath said: (Oct 29, 2013) In the same question what is the probability of taking 2 blue balls?

 Prashant Sinha said: (Dec 7, 2013) In the same question what is the probability of taking 2 blue balls?

 Supriya said: (Mar 25, 2014) How the (7*6)(2*1) has come? Please clear my concept.

 Zuay said: (Apr 23, 2014) No no no! I don't understand are we not suppose to consider probability of not obtaining blue balls here? then its 5/7.

 Sundry said: (May 16, 2014) From the explanation, where did 6 come from? why did you multiply 7 by 6? I did not understand that one.

 Bhavna said: (May 21, 2014) I didn't understand the method which sagar did? can anybody make it clear!

 Hari said: (Aug 19, 2014) Supriya c2 means multiply with number below it. Suppose 2c2= 2*1.

 John said: (Aug 31, 2014) Probability of occurrence of blue ball. 1 blue AND 1 blue OR, 1 red/green AND 1 blue OR, 1 blue AND 1 red/green. (2/7 * 1/6) + (5/7 * 2/6) + (2/7 * 5/6) = 11/21. Probability of non-occurrence of blue ball = 1-(11/21) = (10/21) ANS.

 Sarah said: (Sep 1, 2014) The formula for ncr = n!/(n-r)!*r! if you submit values in this you can get the answer.

 Chi said: (Sep 21, 2014) What is the formula for this question?

 Nagen said: (Dec 21, 2014) Whether sum of probability of 1 blue plus 2 blue is not opposite of none blue.

 Shivam said: (Jan 10, 2015) Why are we taking combination and not permutation in this question?

 Anil Pradhan said: (Jan 27, 2015) Simple and clear way is: Probability of getting blue is 2/7. And getting non-blue is 1-2/7 = 5/7. i.e Total probability-Probability of blue ball.

 Prabhat Kumar Mishra said: (Mar 23, 2015) Total number of balls = 7. Therefore first probability is = 5/7and second probability will be 4/6. Hence, total probability = (5/7)*(4/6) = 20/42 = 10/21.

 Arun Kashyap said: (Apr 14, 2015) Answer is simple: 2+3+2 = 7x3(here 3 is no.of 2+3+2) = 21. 2+ 3(here 3 is no.of 2+3+2) = 5x2 = 10. 10/21.

 Mohammad Javed said: (Apr 16, 2015) There is one easy way. No balls should be blue means either it should be red or green which sums to 5 balls. Total balls is 7. Probability of no blue will be = (5/7)*(4/6)=10/21. As we have picked one ball we are reducing Red+Green balls to 4. And total balls to 6.

 Stephen said: (May 27, 2015) Is there an easier way of doing these type of probability word problems?

 Chandani said: (Jun 16, 2015) Please answer this question. What is the number of ways of selecting 3 balls from a bag containing 5 blue and 6 red balls. If the answer is 11C3 please explain what about the cases where 2 are blue and 1 red. This will happen a number of times, but they will all be counted as separate selections. How do we account for the similarity. Please help.

 Sathish Kumar said: (Aug 13, 2015) Total 7 balls. Blue 2 balls. So none of blue is 5 balls : 5c2/7c2. (5x4/2x1)/(7x6/2x1) = 10/21 answer.

 Itachi said: (Aug 17, 2015) Show the answer using conditional probability please!

 Zishan said: (Sep 4, 2015) There are two bags A and B. A contains n white and 2 black balls and B contains 2 white and n black balls. One of the two bags is selected at random and two balls are drawn from it without replacement. If both the balls drawn are white and the probability that the bag A was used to draw the balls is 6/7, find the value of n. Please answer with solution.

 Jitesh Mittal said: (Sep 15, 2015) Cannot we have a different method. It will be like 1- putting out blue balls which is 2/7. To probability will be 1-2/7 = 5/7?

 Saloni said: (Oct 4, 2015) How we can know that we need to use ''C'' formula in sum ?

 Gmbvbgmkgh said: (Oct 28, 2015) Why is E used?

 Shanel said: (Jan 15, 2016) Hey it should be 5/7. Because if we subtract no of blue from total. Then 7-(2+3) log%5 = 7*5*3*5*6 = log 7+5+3+6+1 = log 35 + log 15 + log 30. So if we take log as common the it will be easy i.e = log (35+15+30). = log*80 = 80 log the 80-75 = 5. There fore the possibility is 5/7.

 Kenneth said: (Feb 11, 2016) Please clear me on how 6 come in?

 Faffy said: (Apr 12, 2016) What if we use the tree diagrams because the way you answer is kindly complicated.

 Rizwan said: (Apr 13, 2016) Why not the answer is 5/7?

 Burhan said: (Jun 16, 2016) Could anybody tell me what is that c means in 7c2?

 Somesh Saurabh said: (Jul 1, 2016) 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.

 Roma said: (Aug 3, 2016) You are right Saurabh, in my way the method 1 is correct.

 Amor said: (Sep 14, 2016) Agree with you @Roma.

 Manoj said: (Sep 26, 2016) 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.

 Namra Shah said: (Dec 7, 2016) Not understanding, please Explain me this method.

 Manisha said: (Dec 12, 2016) 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

 Pooja said: (Dec 17, 2016) 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

 Saravana said: (Dec 26, 2016) What is the need for finding the first step? Please Explain.

 Shah Rukh said: (Jan 8, 2017) @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.

 Ashutosh said: (Jan 13, 2017) Why not the answer is 5/7?

 Abic said: (Jan 23, 2017) The probability of getting blue is 2/7. If we sub 1-2/7 the answer is 5/7 why it is inaccurate? Please explain me.

 Sathya said: (Feb 9, 2017) Here, using combination formulae is good so the answer is 11/21 is correct. Solution:. First possibility 3 balls 7c2 = 21. Second possibility is 5c2 = 11. So that the answer is=11/21.

 Laura said: (Apr 12, 2017) How 7x6? where did you get the 6 from? Please explain it in detail.

 Precious said: (May 5, 2017) 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.

 Sisis said: (May 26, 2017) The Probability of picking 1st ball is 5/7 and second is 4/6 final probability is the product 20/42=10/21.

 Kkp said: (Jun 3, 2017) No answer is none of these. If same colour balls are identical. total possibilities are rr, rg, rb, gg, gb, bb. We don't want bb, gb, rb. So, probability is 3/6 = 1/2.

 Sampathraj said: (Jun 15, 2017) 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].

 Sampathraj said: (Jun 15, 2017) 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].

 Anand Nithi said: (Jul 7, 2017) R2 ,G3, B2. Total = 2+3+2=7. probability = 2/7 *3/7 * 2/7 = 12/21. either R or G ..but not B. probability of B = 2/21. So, (12/21) - (2/21) = 10/21. Is this correct?

 Deepali said: (Jul 10, 2017) I understand the explained method but 7C2 means factorial of 7 so there should be 7*6*5*4*3*2*1/2*1. Wwhy in explanation they took only 7*6. Please explain it.

 Bikash said: (Aug 2, 2017) Can you please explain in detail what is mean by C and all?

 Suyash said: (Sep 13, 2017) Simply, say 5/7 * 4/6. = 10/21.

 Aman said: (Nov 19, 2017) @Jaani. What is the probability of picking 2 blue ball then what is the probability?

 Tenzin said: (Mar 25, 2018) It's answers is 5/7. 1 1,1 2,2 1,3 1,3 2, Five ways that we get the ball without getting a blue ball, So it should be 5/7.

 Pragya said: (Apr 4, 2018) C2 stands for what? Please explain me.

 Trini said: (Apr 11, 2018) 7c2 -> 7!/(7-2)! 2! which is 7*6*5!/5!2! = 7*6/2*1 = 21.

 Ahmed said: (Jun 27, 2018) Simply, find the probability of getting 2 balls blue then subtract from 1.

 Ritu Verma said: (Jul 21, 2018) Prob of 1st ball not blue * prob of 2nd ball not blue. = (5/7)*(4/6), = 20/42, = 10/21.

 Shashikant Sahu said: (Aug 8, 2018) This is a very simple technique. 1st ball not to be blue is = (1-2/7)=5/7. Now next ball not to be blue is = (1-2/6)=4/6. *(causes 1 ball has already taken). Multiple both; = 5/7 * 4/6. = 10/21.

 Harry said: (Aug 8, 2018) 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.

 Manish Kumar Gupta said: (Sep 11, 2018) How come this 2c2/7c2?

 Arnika said: (Oct 11, 2018) What is C and S please explain?

 Christiano Ronaldo said: (Nov 12, 2018) I didn't understand this. Please, anyone, help me to get it.

 Magnus The Great said: (Nov 30, 2018) @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.

 Hemalatha said: (May 9, 2019) @Pranav. The answer is 62.

 Tharini said: (Jul 3, 2019) 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.