# 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.

(5)

Kigele yazidi said:
7 months ago

Total number of balls = 8 + 7 + 6,

=21.

Probability not red or green= number of events/sample space.

So, 7÷21 = 1/3.

=21.

Probability not red or green= number of events/sample space.

So, 7÷21 = 1/3.

(2)

Boopathy said:
6 years ago

Probability of selection Red or Green= 8+6=14 / 21 = 2/3.

Then how 1/3?

Then how 1/3?

(1)

Tushar dhok said:
7 years ago

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.

E= event that the ball is neither red nor green.

= event that the ball is red.

n(E) = 8.

P(E) = 8/21.

(1)

Romeo said:
1 year ago

Thanks all for explaining.

(1)

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.

Rohit said:
5 years ago

How n(E) =7?

Ravindra said:
5 years ago

How can p(e) =7?

Because neither red nor green clearly means its blue.

Because neither red nor green clearly means its blue.

Anitha said:
5 years ago

Total no.of balls =8+7+6=21,

Neither red nor green,

I.e.,we need to take 8,

Therefore p(e)=8/21.

Neither red nor green,

I.e.,we need to take 8,

Therefore p(e)=8/21.

Tharunteja said:
5 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.

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