### Discussion :: Probability - General Questions (Q.No.3)

Monika said: (Mar 11, 2011) | |

HOW n(E)=7? |

Rajeev R T said: (Apr 25, 2011) | |

Event that the ball is neither red nor green, then you are left out with 7 blue balls, and out 7 blue balls you need one ball. Then it is 7C1 ways... i.e = 7 ways. |

Vasu said: (Jul 20, 2011) | |

How the n(e) come as 7? |

Sonu said: (Aug 6, 2011) | |

If at least one is blue means what is the answer? |

Nandini said: (Aug 22, 2011) | |

How n(E) =7? |

S.Sakthivel said: (Aug 30, 2011) | |

Because neither which is indicates that. So that the possibility for the red and green is not considered. Thus simultaneously the blue ball is taken under consideration. |

Sravani said: (Nov 12, 2011) | |

n(E) =7 because there are 7 blue balls. |

Aruna said: (Dec 22, 2011) | |

Neither red nor blue. Then why we are take 7 blue balls only. Can't understand. Please clear my doubt. |

Sahi said: (Dec 30, 2011) | |

Hi aruna I think you didn't read the question properly it is clearly mentioned that neither red nor green, so the remaining are blue. Hence we hve to consider blue balls. Hope you got it :). |

S.Ranga Reddy said: (Feb 25, 2012) | |

Hear there are 1 randomly selected means green 8 blue 6 average of 2colors so 8+6=14 14/2=7 |

Sree said: (Mar 6, 2012) | |

There is a die with 10 faces. It is not known that fair or not. 2 captains want to toss die for batting selection. What is the possible solution among the following? a) If no. is odd it is head, if no. is even it is tail b) If no. is odd it is tail, if no. is even it is head c) Toss a die until all the 10 digits appear on top face. And if first no. in the sequence is odd then consider it as tail. If it is even consider it as head. what is the exact solution |

Hina said: (May 10, 2012) | |

What is the meaning that die is fair or not? |

Angamuthu Suresh said: (Oct 17, 2012) | |

Total number of balls = (8 + 7 + 6) = 21. Let E = event that the ball drawn is neither red nor green =event that the ball drawn is red. Therefore, n(E) = 8. P(E) = 8/21. |

Farah said: (Dec 5, 2012) | |

Please all students read question carefully what he ask in question. He said that 8 red, 6 green, 7 blue total=21 probability ? He said neither from red and green so only one option is 7 blue. 7c1 which is n(a)=7 and n(S)=21c1=21 Formula is prob. p(A)=n(A)/n(S) 7/21 = 1/3. |

Muhammad Imran Tariq said: (Dec 24, 2012) | |

Sample= 8+7+6=21. Probability of selection Red and Green= 8+6=14 / 21 = 2/3. Not selection of Red and Green. Simple subtracts 1 from 2/3. 1 - 2/3= 1/3 (Answer). |

Mangesh Modiraj said: (Jul 8, 2013) | |

@Nandini & @Monica. In example clearly mentioned that the ball ins neither red nor green. i.e blue. We have 7 blue balls. Therefore n(E) = 7. |

Bibekpangeni said: (Feb 4, 2014) | |

What is the meaning of probability? |

Durga said: (Apr 26, 2014) | |

Total balls are - 21 So the probability of getting one ball is -21c1 = 21. That one ball is neither red nor green means blue ball. Posing one blue ball from 7 blue balls is 7c1 = 7. Our required ans=7/21=1/3. |

Srinu said: (Feb 7, 2015) | |

Total balls-21. 7C1 = 7. 7/21. |

Otuoboizy said: (Oct 23, 2015) | |

Total numbers of balls is 21 and cl is 7. |

Knowledgeispower said: (Feb 6, 2016) | |

Using the same problem, solve for the probability of pulling a RED ball on 2 draws without replacing the ball after each draw. |

Faffy said: (Apr 12, 2016) | |

Does this mean that we are excluding green and red? |

Obiano Kingseley said: (Jun 1, 2016) | |

A bag contains 5yellow balls, 3green balls, and 2red balls, a ball is chosen at random from the bag. What is the probability that the chosen ball is yellow? Can anyone answer this? |

Ramachandran Nichite said: (Aug 15, 2016) | |

Hi @Obiano. Total ball = 5 + 3 + 2 = 10. Choose yellow ball = 5/10. |

Jaiprakash said: (Nov 6, 2016) | |

@Obiano. The answer is 1/2. |

Tushar Dhok said: (May 2, 2017) | |

Total no.of ball 8, 7, 6 = 21. E= event that the ball is neither red nor green. = event that the ball is red. n(E) = 8. P(E) = 8/21. |

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