Discussion :: Probability - General Questions (Q.No.10)
n(E) = 15.| Ram Reddy said: (Dec 2, 2010) | |
| Please tell me any other method. | |
| Srinivas said: (Dec 15, 2010) | |
| Hi sir, i am little bit poor in maths, what is the meaning of prime numbers... { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) } Why do choose those no only ? Why don't you choose (6,6) (2,2)n etc |
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| Nachiket said: (Jan 27, 2011) | |
| Prime number is a number which is divisible by 1 and that number only. | |
| Debai said: (Jan 28, 2011) | |
| Hi RAM, if you tossed 2 dice then you get total 6x6 ie 36 possible cases (equallay likely cases) the above mentioned 15points are prime nos (2, 3, 5, 7, 11, 13, 17, 19. ) if you add any one out of 15 get prime no. That's all. | |
| Sujatha said: (Jul 25, 2011) | |
| Could you please tel me any short methods for Probability? | |
| Harshini said: (Sep 14, 2011) | |
| Please tel me where to use permutation and where to use combinations. ? | |
| Prasad said: (Jan 16, 2012) | |
| Hey why we cant take (1,3),(2,2,),(3,1)etc we can take others also na? So please solve this in easy way. |
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| Anil Singh said: (Jan 18, 2012) | |
| Prime number divisible by 1 and that nimber only but in the case (1,4), (1,6) etc divided by other number like 1,2,3 Please explain. |
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| Olamikan Adesanmi said: (Apr 13, 2012) | |
| No wot they mean is that the sum (total) is a prime number. So as for (1, 4) , (1, 6) , the sum are 5 and 7 which are prime numbers. | |
| Rakesh said: (Jul 25, 2012) | |
| Why we not chosen (2, 2) because it also prime number and it divisible by only 2? | |
| Ramya said: (Nov 18, 2012) | |
| They have left out 6, 3 and 3, 6 then 5, 6 and 6, 5. These also should be included then the total probability becomes 1/2. | |
| Ravindra said: (Nov 28, 2012) | |
| @Harshini. Permutations are used for arrangement where as combinations are used for selection. Ex: Arrange the name xyz then per is used. Where as, Select one among this xyz comb is used. |
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| Siva said: (Feb 16, 2013) | |
| Prime number means the we gets only one time. i.e., 1*2=2 only once we get 2. But for 4 1*4=4, 2*2=4 now we get 4 for 2 times in-terms of multiplication. | |
| Ganesh Sang said: (Feb 22, 2013) | |
| Please in every example we need to find samples or experiment samples or there is any shortcut for easily getting number of prime number or sum of two numbers as a prime ? | |
| Suvendu said: (Feb 25, 2013) | |
| As the question says sum should be a prime number. So sum must be 2, 3, 5, 7, 11, So probability should be (1, 1) (1, 2) (2, 1) (1, 4) (4, 1) (3, 4) (4, 3) (5, 6) (6, 5). So probability= (9/36) =1/4. |
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| Akshay said: (Apr 26, 2013) | |
| Why don't we consider (1, 1) two times. Because it will be the outcome of both the dices. | |
| Sowmya said: (Jul 14, 2013) | |
| The sample space is a set, so repeated event (1, 1) is taken only once. | |
| Amit said: (Aug 23, 2013) | |
| Why (1, 1) don't consider two times while sets like (1, 2) n (2, 1) is taken two times? | |
| Sikkanthar said: (Aug 28, 2013) | |
| Finally I found the answer of this question. It is very simple. When two dice are tossed the total number of possibilities are 36. Which means 6*6 =36. They are asked to find out the prime number. We know what is prime number, the number is divisible by the same number except 1. In this concept the possibilities are (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),(5, 2), (5, 6), (6, 1), (6, 5). from {(1,1), (1,2)... (6,6)}. 1+1=2, 1+2=3, 1+4=5, 1+6=7 the same way you can get the indivisible numbers expect the same number. So answer is 15/36 which is equal to 5/12 is the correct answer :-). |
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| Vrushang said: (Nov 21, 2013) | |
| Hey if you take 5, 6 and 6, 5 as a different pair then why you don't take (1, 1) and (1, 1) as different pairs because it is first time 1 and second time 1. | |
| Jhansi Sri said: (Feb 12, 2014) | |
| @Vrushang. In combination(1, 1) are same, in permutations order is impt ex: (1, 2),(2, 1). Are same selection in combs we first 1 or 2, 2 or 1 is same comb. But in permutations the arrangement is imp 1, 2 it is different from 2, 1 first 1 is set in the row next second place 2 is set, first 2 is set in the row, next 1 is set in the row. So I think you are understand. |
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| Mayur said: (Apr 7, 2014) | |
| Prime number is nothing but a number which is divided by 1 or itself. For example 5 is divided by itself or only 1. So its called a prime number. 6 is not called a prime no because of it can divide by 2, 3. I hope now clear cut. | |
| Harshal said: (Jul 12, 2014) | |
| 1 is prime no or composite no? | |
| Bumba said: (Jul 18, 2014) | |
| If we can chose (1,2) and (2,1) (2,3) and (3,2). Then why we can't chose (1,1) both time. Not only that case we also chose (1,1) as a double event space. Please reply. |
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| Tuhin said: (Sep 4, 2014) | |
| We have prime numbers between 2 To 12 as 2,3,5,7,11. There is 1 Way to get sum as 2, 2 ways to get sum 3, 4 To get 5, 6 Ways to get 7 And 2 Ways to get 11. So total of 15, therefore prob = 15/36. |
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| Sujit said: (Sep 8, 2014) | |
| Don't you think we should not count (1,2) and (2,1) as 2 it must counted as 1 only. | |
| Amandeep said: (Oct 10, 2014) | |
| How we got 15 point please anyone can explain? | |
| Pankaj said: (Nov 7, 2014) | |
| No detail about the Dice's aren't they Identical (0,0). | |
| Aniket said: (Dec 20, 2014) | |
| The total no of unique combinations on throw of 2 dice is 36: i.e. {(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(1,4),(4,1),(2,3),(3,2),(1,5),(5,1),(2,4),(4,2),(3,3),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(2,6),(6,2),(3,5),(5,3),(4,4),(3,6),(6,3),(4,5),(5,4),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)}. Out of these pairs, how many give us a prime number sum = 15. i.e. {(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(5,6),(6,5)}. So, probability = 15/36 = 5/12. |
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| Saranya said: (Jan 2, 2015) | |
| Why 4 kings are taken? | |
| Sasikala said: (Feb 1, 2015) | |
| Could any one of you explain me in simplistic processes? | |
| Indrajit said: (Nov 30, 2015) | |
| Hey @Prasad, Here is the clarification of your doubt. You have to choose those number of pairs whose sum is a prime number. If you choose (1, 3), (2, 2) & (3, 1) then their sum will be 4 which is not a prime number. Thanks! |
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| Osei James said: (Apr 26, 2016) | |
| If two numbers are selected at random, one after the other with replacement from the set A = (5, 6, 7, 8, 9). Find the probability of selecting at least one prime number. Can anyone help me to solve this problem. |
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| Azly said: (Jun 19, 2016) | |
| If we draw this in a cartesian plan we can easily find the answer. Take d axis and Y axis. Number from 1 to 6 in each axis. The find the sum of the numbers using plan. | |
| Shikha Patel said: (Jul 19, 2016) | |
| Thanks @Azly. By using axis method, we can easily solve the question like this. |
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| Vipin said: (Jul 30, 2016) | |
| Why we take (1,1) while prime number started from 3, 4, 5, 7? | |
| Abhilasha said: (Aug 6, 2016) | |
| @Vipin. 2 is also a prime number that's the reason we take (1,1). |
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| Nitin Rajput said: (Sep 3, 2016) | |
| Thanks sir ,your are brilliant because you give me a platform to improve our objective paper skills.. | |
| Kiran said: (Oct 13, 2016) | |
| You are considering (1, 1) is a prime set and why not you consider (2, 2), (3, 3), (5, 5) as a prime set? | |
| Nilesh Patil said: (Oct 20, 2016) | |
| (1,1), (2,1), (2,3), (3,2), (4,1), (1,4), (4,3), (3,4), (5,2), (2,5), (6,1), (1,6), (6,3), (3,6), (6,5),(5,6) If(1,1) is considered then n(E) becomes 16 then, what's wrong in above? |
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| Lionel said: (Nov 12, 2016) | |
| Assume that two independent random samples of size 100 each is taken from a population that has a variance of 36. What is the probability that the difference in the sample means is more than 2? Please give me the solution. |
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| Tahir Amin said: (Dec 21, 2016) | |
| Please, Solve anyone this question. A committee of 3 is to be selected from a group of 5 girls and 3 boys. What is the probability that two boys are selected? | |
| Sowmya said: (Dec 21, 2016) | |
| Is total score nothing but the sum? | |
| Karma said: (Apr 15, 2017) | |
| Here, I used nCr. But its not working, why? Can anyone explain me. |
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| Supriya said: (Sep 18, 2017) | |
| Can anyone explain the simplest method to find numarator? | |
| Mirekua said: (Oct 17, 2017) | |
| Please, from where E={(1,1),(1,2)...} is came? Please explain. | |
| Mahendra said: (Dec 28, 2017) | |
| Why we enter prime numbers sum as 15? Please explain the step. | |
| Suhail Abdul Rehman Chougule said: (Jan 23, 2018) | |
| I am sure the answer is 1/6 for this as this event mentioned has something missing. E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) } E(correct) = { (1,1), (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) } (1,1) will have a probability 2/12 rather than 1/12 un-like other pairs. If you note (2,1) occurs and (1,2) also occurs and hence (1,1) due to dice A and additionally (1,1) due to dice B and hence the probability of this event is doubled. |
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| Abs Rasel said: (Feb 25, 2018) | |
| Prime numbers 2={1,1} 3={1,2};{2,1} 5={1,4};{4,1};{2,3};{3,2} 7={1,6};{6,1};{2,5};{5,2};{3,4};{4,3} 11={5,6};{6,5} Total 15 possible outcomes. So, Probability is =15/36=5/12. |
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| Durba said: (Jun 19, 2018) | |
| The sum of (1, 4) (1, 6) is not prime. | |
| Sirisha said: (Dec 17, 2018) | |
| 1+1=2 here 2 is not a prime number then please explain why (1, 1) used here? Explain, please. |
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| Williams said: (May 12, 2019) | |
| In the answer, why didn't they count (1,1) twice as it is possible that the number 1 might come on either of the dices, as they calculated (1,2) & (2,1) | |
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