# 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 2 of 4.
Rajeev R T said:
1 decade ago

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.

Then it is 7C1 ways... i.e = 7 ways.

Muhammad Imran Tariq said:
1 decade ago

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).

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).

Kigele yazidi said:
10 months ago

Total number of balls = 8 + 7 + 6,

=21.

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

So, 7รท21 = 1/3.

=21.

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

So, 7รท21 = 1/3.

(3)

Mangesh Modiraj said:
1 decade ago

@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.

In example clearly mentioned that the ball ins neither red nor green.

i.e blue.

We have 7 blue balls.

Therefore n(E) = 7.

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)

Knowledgeispower said:
8 years ago

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

Aruna said:
1 decade ago

Neither red nor blue.

Then why we are take 7 blue balls only.

Can't understand.

Please clear my doubt.

Then why we are take 7 blue balls only.

Can't understand.

Please clear my doubt.

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.

S.Ranga reddy said:
1 decade ago

Hear there are 1 randomly selected means

green 8 blue 6

average of 2colors

so 8+6=14

14/2=7

green 8 blue 6

average of 2colors

so 8+6=14

14/2=7

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)

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