# Aptitude - Probability - Discussion

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

3.

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Answer: Option

Explanation:

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 blue. |

*n*(E) = 7.

P(E) = | n(E) |
= | 7 | = | 1 | . |

n(S) |
21 | 3 |

Discussion:

36 comments Page 1 of 4.
Chris said:
2 years ago

@All.

Here, You can solve it in two ways.

Method 1:

Total number of balls = 7+8+6= 21.

Neither red nor green, that means it is blue,

Number of blue = 7,

Probability of blue = number of blue ballsÃ· total number of balls.

Probability of blue balls= 7/ 21.

1/3.

Method 2:

Total number of balls = 21.

Number of Red = 8.

Prob of selecting red = 8/21 or.

Number of Green = 6.

Prob of selecting green = 6/21 or 2/7.

Neither Red nor Green = 1-(prob of red+ prob of green).

1 - (8/21 + 6/21).

1 - 14/21 = 7/21 or 1/3.

Here, You can solve it in two ways.

Method 1:

Total number of balls = 7+8+6= 21.

Neither red nor green, that means it is blue,

Number of blue = 7,

Probability of blue = number of blue ballsÃ· total number of balls.

Probability of blue balls= 7/ 21.

1/3.

Method 2:

Total number of balls = 21.

Number of Red = 8.

Prob of selecting red = 8/21 or.

Number of Green = 6.

Prob of selecting green = 6/21 or 2/7.

Neither Red nor Green = 1-(prob of red+ prob of green).

1 - (8/21 + 6/21).

1 - 14/21 = 7/21 or 1/3.

(10)

Sree said:
1 decade ago

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

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

Tharunteja said:
6 years ago

Guys it may be a wrong answer.

The actual correct answer;

Total no.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 (given neither red nor green) make a note this point.

therefore p(E) = 8/21.

The actual correct answer;

Total no.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 (given neither red nor green) make a note this point.

therefore p(E) = 8/21.

Farah said:
1 decade ago

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.

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.

Durga said:
1 decade ago

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.

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.

Sahi said:
1 decade ago

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 :).

Obiano kingseley said:
8 years ago

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?

Can anyone answer this?

Shiv said:
3 years ago

Neither red''ï¸ nor green'' means (avoiding red and green) so we left out with blue means the one ball we have to pick up should be blue colour ball so.

P(E) = 7/21 = 1/3.

P(E) = 7/21 = 1/3.

S.Sakthivel said:
1 decade ago

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.

Angamuthu SURESH said:
1 decade ago

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.

= 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.

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