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?
1
3
3
4
7
19
8
21
9
21
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.

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.
(1)

Jaiprakash said:   8 years ago
@Obiano.

The answer is 1/2.

Ramachandran Nichite said:   8 years ago
Hi @Obiano.

Total ball = 5 + 3 + 2 = 10.

Choose yellow ball = 5/10.

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?

Faffy said:   8 years ago
Does this mean that we are excluding green and red?

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.

Otuoboizy said:   9 years ago
Total numbers of balls is 21 and cl is 7.

SRINU said:   9 years ago
Total balls-21.
7C1 = 7.
7/21.

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.

Bibekpangeni said:   1 decade ago
What is the meaning of probability?


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