# 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 3 of 4.
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.

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).

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.

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.

Hina said:
1 decade ago

What is the meaning that die is fair or not?

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

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

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 :).

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.

Sravani said:
1 decade ago

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

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