A Box Contains Six Balls, 2 White, 2 Black, 2 Red What Is The Sample Space For One Random Draw?
| 2. | A handbag contains 2 red, 3 dark-green and ii bluish balls. Ii balls are drawn at random. What is the probability that none of the balls drawn is blue? | ||||||||||||||||||||||
| Answer: Selection A Explanation: Full number of balls = (2 + 3 + 2) = 7. Let Due south be the sample space.
Allow E = Issue of drawing 2 balls, none of which is blue.
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n(E)| 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 right answer and short way. | |
| Kk585 said: (December 24, 2010) | |
| Probability that the 2 balls r blueish ----> 2c2/7c2 = 1/21 Probability that none of ball is blueish ---> 1-(1/21) = 20/21 Why is my answer wrong delight 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 vi and by 4? | |
| Rose Were said: (May 19, 2011) | |
| How did the multiplications come about? please respond me. | |
| Subhash said: (Jun 3, 2011) | |
| What Kk585 said is not right. Becuase, here we should consider no ball is bluish. pro.of 2 bluish balls is (2c2/7c2), pro.of 1blue ball and another one 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 one b and 1 red is ((2c1/7c1)*(2c1/6c1))*2. Therefore total is eleven/21. Hence ans is 1-(11/21) is x/21. | |
| Ekta said: (Aug 8, 2011) | |
| Please clarify it which 1 having best respond of the solution. | |
| Akash said: (Sep 26, 2011) | |
| Can we practise information technology by this way? 3c2+2c2+3c1 2c1 I have washed information technology this way. 3C2 ways that the both balls would be green 2C2 both would be carmine and 3C1 and 2C1 because both the balls tin can be red and green. Isn't the solution correct? | |
| Shiba said: (Sep 27, 2011) | |
| n(E) = Number of ways of drawing ii balls out of (two + three) balls. ? My question is why are you finding it only from 5 assurance why not 7 balls. | |
| Srija said: (November 5, 2011) | |
| You are right akash! | |
| Dileep said: (Feb 14, 2012) | |
| The fashion akash did is exact. It takes into account all the probabilities | |
| Jay said: (Mar 2, 2012) | |
| (7 x six) (2 x i) Please explan it. | |
| Neha said: (Mar 29, 2012) | |
| What is 3c2, 2c2 etc? could anyone please explain this whole sum according to a form 7 student. | |
| Pranav said: (May 10, 2012) | |
| A handbag contains iii crimson, 2 greenish and 5 blue balls. 4 balls are fatigued at random. What is the probability that ane green ball? | |
| Madhu said: (Jul viii, 2012) | |
| Becuase, here nosotros should consider no brawl is bluish. pro.of 2 bluish balls is (2c2/7c2), pro.of 1blue ball and another 1 green brawl is((2c1/7c1)*(3c1/6c1))*2. Here ii means b+g or thou+b means will occur. In the similar manner ,pro. of 1 b and one red is ((2c1/7c1)*(2c1/6c1))*ii. | |
| Nishant said: (Aug 13, 2012) | |
| Can you please explain me why we select Number of ways of drawing two balls out of (two + 3) balls instead of there having total no.of vii assurance, . And so prob. Of drawing 2 balls must be out of 7 i.due east. (2+iii+two). | |
| Krish said: (Aug 17, 2012) | |
| Because, here nosotros should consider no ball is blue. pro.of two bluish assurance is (2c2/7c2), pro.of 1blue ball and another 1 greenish ball is((2c1/7c1)*(3c1/6c1))*2. Here 2 means b+m or g+b means will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*two. But the question says NO Bluish BALL..why are we considering ane blue and 1 greenish ...or ane redor 1 blue..? | |
| Krish said: (Aug 17, 2012) | |
| I am not geting wats wrong in this...please explain.... Probability that the 2 balls r bluish ----> 2c2/7c2 = 1/21 Probability that none of ball is blue ---> 1-(1/21) = xx/21 Why is my reply wrong please explain? p(a)+p(a(bar)) = 1 ....... | |
| Jaani! said: (Aug 23, 2012) | |
| @krish, Cuz 2C2 ways picking those 2 blue assurance out of those two bluish balls simply! Bt d ques says "None of them bluish" bt one blue and one of any other colour cant b included too! Thats y we do that of picking upwardly only green and reddish.. (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) =north (E) /n (due south) =eleven/21. Simply we have to discover hither is none of the 2 ball is blue so we have to find. P (~E) =1-P (East) =10/21. | |
| Piyul said: (Nov 22, 2012) | |
| Is there whatever other way to understand basic concept of probability? | |
| Farah said: (Dec v, 2012) | |
| Formula is probability is p(A) = northward(A)/north(southward), then we find n(s) = 7C2 = 21. Next we find n(a) = 5C2 = 10 these amounts put in formula p(A)=north(A)/n(Southward) Ans is 10/21 Nosotros 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 Ruddy iii Green ii Blueish Assurance i.e. 5 Non blue assurance and 2 Bluish 1st ball tin be picked from those v non blue balls and then that none of them is blue. Hence northward(E)=v & due north(Due south)= 7 The next ball will be picked from remaining 6 balls and four Not-Blue balls(Since one is already picked) Hence Probability(second picked brawl not blue) = 4/6 Hence | |
| Kavita Kant said: (Mar xi, 2013) | |
| Is there whatsoever fashion to determine that when nosotros take combination to find probability and when nosotros 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) = vii. Let S be the sample space. And then, north(S) = Number of ways of drawing 2 balls out of seven A: Event none of the balls draw is bluish. northward(A): number of possible ways = 2C1*3C1+2C2*3C0+3C2*2C0 P(A)=n(A)/due north(S)=10/21=1/2. | |
| Sandeep said: (Jul 31, 2013) | |
| If you are not aware of combinations like 7C2 and then give names to the balls similar r1, r2, g1, g2, g3, b1 and b2. So if you take ii balls out of the bag then if nosotros consider r1 comes then total combination would be r1r2, r1g1, r1g2.. and and so on. That means, for r1 there are 6 combinations. So we can conclude post-obit tabular array. Ball combination So total number of combination = six+5+4+3+2+1 = 21. So combination without using blue brawl = 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 two blueish balls? | |
| Prashant Sinha said: (Dec 7, 2013) | |
| In the same question what is the probability of taking 2 blueish balls? | |
| Supriya said: (Mar 25, 2014) | |
| How the (7*six)(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/seven. | |
| Sundry said: (May sixteen, 2014) | |
| From the explanation, where did 6 come from? why did you multiply 7 past 6? I did not understand that one. | |
| Bhavna said: (May 21, 2014) | |
| I didn't understand the method which sagar did? can everyone get in articulate! | |
| Hari said: (Aug 19, 2014) | |
| Supriya c2 ways multiply with number below information technology. Suppose 2c2= 2*one. | |
| John said: (Aug 31, 2014) | |
| Probability of occurrence of blue ball. 1 blueish AND 1 blue OR, (2/7 * 1/half-dozen) + (v/7 * 2/6) + (2/7 * five/6) = eleven/21. Probability of not-occurrence of bluish brawl = 1-(11/21) = (ten/21) ANS. | |
| Sarah said: (Sep 1, 2014) | |
| The formula for ncr = north!/(n-r)!*r! if you submit values in this you tin can become the answer. | |
| Chi said: (Sep 21, 2014) | |
| What is the formula for this question? | |
| Nagen said: (Dec 21, 2014) | |
| Whether sum of probability of i blue plus two blue is not opposite of none bluish. | |
| Shivam said: (January 10, 2015) | |
| Why are we taking combination and not permutation in this question? | |
| Anil Pradhan said: (January 27, 2015) | |
| Simple and clear way is: Probability of getting blue is ii/7. And getting non-blue is 1-ii/7 = v/7. i.due east Total probability-Probability of blue ball. | |
| Prabhat Kumar Mishra said: (Mar 23, 2015) | |
| Total number of balls = seven. Therefore first probability is = 5/7and 2nd probability volition exist 4/6. Hence, total probability = (five/vii)*(4/6) = xx/42 = 10/21. | |
| Arun Kashyap said: (Apr fourteen, 2015) | |
| Answer is elementary: 2+3+2 = 7x3(here 3 is no.of 2+3+2) = 21. 2+ three(here 3 is no.of two+3+2) = 5x2 = 10. 10/21. | |
| Mohammad Javed said: (Apr 16, 2015) | |
| At that place is one easy way. No balls should exist blue means either information technology should be red or light-green which sums to 5 balls. Total balls is seven. Probability of no blue volition exist = (5/vii)*(4/half dozen)=10/21. As nosotros take picked one ball we are reducing Reddish+Green balls to 4. And total balls to half-dozen. | |
| Stephen said: (May 27, 2015) | |
| Is at that place an easier way of doing these type of probability give-and-take problems? | |
| Chandani said: (Jun 16, 2015) | |
| Please respond this question. What is the number of ways of selecting iii balls from a bag containing v blue and 6 red assurance. If the answer is 11C3 please explain what almost the cases where 2 are bluish and 1 cerise. This will happen a number of times, only they will all be counted as carve up selections. How do we account for the similarity. Please aid. | |
| Sathish Kumar said: (Aug 13, 2015) | |
| Total 7 balls. Blue ii balls. And so none of bluish is five assurance : 5c2/7c2. (5x4/2x1)/(7x6/2x1) = 10/21 answer. | |
| Itachi said: (Aug 17, 2015) | |
| Show the respond using provisional probability delight! | |
| Zishan said: (Sep four, 2015) | |
| At that place are two bags A and B. A contains northward white and 2 black balls and B contains ii white and northward black balls. One of the two numberless is selected at random and 2 balls are fatigued from information technology without replacement. If both the balls drawn are white and the probability that the bag A was used to depict the assurance is six/vii, detect the value of n. Please respond with solution. | |
| Jitesh Mittal said: (Sep 15, 2015) | |
| Cannot we have a dissimilar method. Information technology will be like ane- putting out blueish balls which is ii/7. To probability will be 1-2/7 = five/7? | |
| Saloni said: (October iv, 2015) | |
| How nosotros can know that we need to employ ''C'' formula in sum ? | |
| Gmbvbgmkgh said: (Oct 28, 2015) | |
| Why is E used? | |
| Shanel said: (Jan fifteen, 2016) | |
| Hey it should be 5/7. Because if nosotros decrease no of blue from full. So 7-(2+3) log%5 = 7*five*3*5*6 = log 7+5+3+half-dozen+1 = log 35 + log xv + log 30. So if we take log as mutual the it will exist easy i.e = log (35+15+30). = log*80 = 80 log the 80-75 = 5. There fore the possibility is 5/7. | |
| Kenneth said: (February 11, 2016) | |
| Please clear me on how vi come in? | |
| Faffy said: (April 12, 2016) | |
| What if we use the tree diagrams because the way you respond is kindly complicated. | |
| Rizwan said: (Apr 13, 2016) | |
| Why not the answer is 5/vii? | |
| Burhan said: (Jun xvi, 2016) | |
| Could everyone tell me what is that c ways in 7c2? | |
| Somesh Saurabh said: (Jul 1, 2016) | |
| See in that location are ii means to solve this question: METHOD 1: Probability of getting no blueish balls in 2 draws = probability of getting 2 other balls in two draws out of any other v balls. i.due east., select two balls from the remaining 5 balls in 5C2 means = n(E), At present the sample space is n(Southward) = selection of 2 balls out 7b balls = 7C2, So, required probability= n(E)/n(Due south), => 5C2/7C2 = [5!/(2!*three!)]/[vii!/(two!*5!)] => x/21 is the reply. METHOD 2: Probability of not getting any blue balls in ii draws = ane - (probability of getting ane blue ball and 1 other ball + probability of getting 2 bluish balls) And so, Pick of 1 blue ball and one other ball = 2C1 * 5C1 = 10 ways, Option of 2 blueish assurance = 2C2 = 1, So, probability = 1 - [{x/7C2} + {one/7C2}] => x/21 answer. | |
| Roma said: (Aug three, 2016) | |
| You are correct Saurabh, in my way the method ane is correct. | |
| Amor said: (Sep 14, 2016) | |
| Concur with you @Roma. | |
| Manoj said: (Sep 26, 2016) | |
| How did y'all obtain 21 sample spaces in the to a higher place-mentioned solution? Tin you mention all possible combination of balls? Ex:{ HH, TT, HT, Thursday } 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 banking concern contains 5white,6red,2green, and 2black assurance ane ball is selected at random from the pocketbook. Find the probability that the selected ball is: 1. White | |
| Pooja said: (December 17, 2016) | |
| Please solve this question: A bag contains 5 white, six red, 2 dark-green and 2 black balls. One ball is selected at random from the bag. Find the probability that the selected brawl is; A) white | |
| Saravana said: (December 26, 2016) | |
| What is the need for finding the first step? Please Explicate. | |
| Shah Rukh said: (Jan 8, 2017) | |
| @Pooja for A. White. Ans = totql assurance = 5+6+two+ii = xv. Totql no. Of outcomes i.e. S=15c1 =15/1 = 15. Issue A = 5c1 = 5. Probability of event A = 5/xv = 1/iii. | |
| Ashutosh said: (Jan 13, 2017) | |
| Why non the answer is five/7? | |
| Abic said: (Jan 23, 2017) | |
| The probability of getting blue is ii/seven. If nosotros sub 1-ii/7 the answer is 5/7 why it is inaccurate? Please explain me. | |
| Sathya said: (February 9, 2017) | |
| Here, using combination formulae is good then the answer is 11/21 is correct. Solution:. Starting time possibility 3 assurance 7c2 = 21. So that the answer is=xi/21. | |
| Laura said: (Apr 12, 2017) | |
| How 7x6? where did yous get the 6 from? Please explicate it in item. | |
| Precious said: (May 5, 2017) | |
| When we say 6C subscript ii, information technology means six * 5 * 4 * 3 * ii * 1 divided by 2 * 1. Then, the right reply to me is 5/vii. To find the probability that none of the blueish ball is drawn, 1 must be subtracted from the probability that the blue ball was chosen at random. i - 2/7. | |
| Sisis said: (May 26, 2017) | |
| The Probability of picking 1st ball is 5/7 and second is 4/half-dozen last probability is the product twenty/42=10/21. | |
| Kkp said: (Jun 3, 2017) | |
| No reply is none of these. If same colour assurance are identical. total possibilities are rr, rg, rb, gg, gb, bb. We don't desire bb, gb, rb. And then, probability is 3/6 = i/2. | |
| Sampathraj said: (Jun fifteen, 2017) | |
| Given criteria - Need the upshot to be either Red Or Green, but not Blueish. So, at first - The Total poswsible combination outcomes(Events) we would get considering Red Or Dark-green(2+3 = five) i.e "5c2" Sample space(All Outcomes) = (Red+Light-green+Blue = 7) ; "7c2 | |
| Sampathraj said: (Jun 15, 2017) | |
| Given criteria - Need the event to exist either Red Or Green, only not Blueish. So, at first - The Total possible combination outcomes(Events) we would become considering Cherry-red Or Light-green(ii+three = five) i.eastward "5c2". Sample space(All Outcomes) = (Scarlet+Green+Blue = 7) ; "7c2 = 5c2 / 7c2, | |
| Anand Nithi said: (Jul seven, 2017) | |
| R2 ,G3, B2. Full = 2+iii+2=7. And so, Is this right? | |
| Deepali said: (Jul 10, 2017) | |
| I empathize the explained method but 7C2 ways factorial of 7 so there should be 7*6*5*4*3*2*1/2*ane. Wwhy in explanation they took only 7*6. Please explain it. | |
| Bikash said: (Aug two, 2017) | |
| Can you please explain in particular what is hateful past C and all? | |
| Suyash said: (Sep thirteen, 2017) | |
| But, say five/7 * 4/half dozen. = 10/21. | |
| Aman said: (Nov 19, 2017) | |
| @Jaani. What is the probability of picking 2 blue ball so what is the probability? | |
| Tenzin said: (Mar 25, 2018) | |
| It's answers is 5/7. 1 1,1 2,two one,iii 1,3 2, Five ways that we go the ball without getting a blue ball, | |
| Pragya said: (Apr iv, 2018) | |
| C2 stands for what? Delight explain me. | |
| Trini said: (Apr 11, 2018) | |
| 7c2 -> 7!/(7-2)! 2! which is seven*half-dozen*5!/5!2! = seven*six/2*i = 21. | |
| Ahmed said: (Jun 27, 2018) | |
| But, find the probability of getting ii balls bluish then subtract from ane. | |
| Ritu Verma said: (Jul 21, 2018) | |
| Prob of 1st ball non bluish * prob of 2nd ball non blue. = (5/7)*(4/6), | |
| Shashikant Sahu said: (Aug 8, 2018) | |
| This is a very simple technique. 1st ball non to be blueish is = (1-two/7)=v/vii. At present next ball non to be blue is = (1-2/6)=4/half dozen. Multiple both; = v/7 * iv/6. | |
| Harry said: (Aug eight, 2018) | |
| Can someone help me with this office of the question: Allow S exist the sample space. Then, due north(S) = Number of ways of drawing ii balls out of 7. I don't understand what 7C2 means or how nosotros get 7x6 or 2x1 and then divided. | |
| Manish Kumar Gupta said: (Sep 11, 2018) | |
| How come up this 2c2/7c2? | |
| Arnika said: (Oct 11, 2018) | |
| What is C and S delight explain? | |
| Christiano Ronaldo said: (Nov 12, 2018) | |
| I didn't understand this. Please, anyone, assist me to get it. | |
| Magnus The Not bad said: (November thirty, 2018) | |
| @ALL. Co-ordinate to me, Notation: (Always solve as *without replacement* unless otherwise stated) Pr(1st Red & 2nd ruddy...RR) = two/7 * ane/6 = 1/21, And then | |
| Hemalatha said: (May 9, 2019) | |
| @Pranav. The answer is 62. | |
| Tharini said: (Jul 3, 2019) | |
| In that location is; 2 Red. 1st ball tin can be picked from those 5 non-blue balls then that none of them is blue. Hence n(E)=v & n(Due south)= seven. The next ball will exist picked from remaining 6 balls and four Not-Blueish balls(Since one is already picked). Hence Probability(2d picked ball not bluish) = 4/vi. Hence, | |
| Summi said: (Jan 27, 2020) | |
| How come seven*6/2*i? please clear me. | |
| Abi said: (Jun iv, 2020) | |
| The reply should be 5/7 right! Because Probability = No.of favourable outcomes/Full outcomes. Since, if both balls r not bluish and then they accept to exist either red(2) or green (3). And then, option D is correct. | |
| Krish said: (Jun 18, 2020) | |
| Using conceptual method rather than going past formula gives a better feel for what is happening. P1 = probability of 1st brawl not blue = 5/7 as v are non-blue. | |
| Naseem Ullah Khan said: (Aug 21, 2020) | |
| Simple explanation, well washed, thanks @Krish. | |
| Gufran said: (Aug 22, 2020) | |
| (2R+3G)c2/7c2. 5c2/7c2. 5 * 2/ii/7 * 6/two. (20/two)/(42/2). 10/21 is the final answer | |
| Sai said: (Aug 26, 2020) | |
| 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 * half-dozen = 42/ii = 21. e = drawing balls none of which is blueish, Therefore p n(e)= 10/21. | |
| Biswajit said: (Aug 29, 2020) | |
| Can anyone tell me if I subtract the probability of getting 2 blue balls i.e 2/seven from ane then why I tin't get the answer? Please explain me. | |
| Kanishk said: (Sep 1, 2020) | |
| @Biswajit. Firstly the probability of drawing 2 blueish balls will be 2/seven*i/6. Promise y'all get information technology now! | |
| Thejo said: (Oct 7, 2020) | |
| We should pick 2 from vii none of the ball fatigued is bluish, then the picked assurance colours should be red and green. How come information technology is ten/21? Here it could be 2C1*3C1 ÷ 7C2 = 6/21 right! Tin any1 tell me why it is wrong? | |
| Srz said: (Oct 18, 2020) | |
| P1st (non blue)= 1-P(blue) = i-2/seven = five/7. P2nd(not blueish)= i-P(blueish) = one-2/6 = iv/half dozen. P1 and P2(not blue)=v/7 x 4/6 = xx/42 = 10/21. | |
| Lyi said: (Nov three, 2020) | |
| The probability of the first ball non beingness blue is 5/seven and the probability of the 2d ball also not existence blue is 4/6 and this is and so considering we have already drawn one ball and at present the total number of balls got reduced to half dozen. At present where the iv comes from? Its because we're bold that the first ball we've drawn is a non-blue and therefore nosotros have i less brawl than 5 to describe from now. I hope it was helpful. | |
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