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.
Knowledgeispower said:
10 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.
Faffy said:
9 years ago
Does this mean that we are excluding green and red?
Obiano kingseley said:
9 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?
Ramachandran Nichite said:
9 years ago
Hi @Obiano.
Total ball = 5 + 3 + 2 = 10.
Choose yellow ball = 5/10.
Total ball = 5 + 3 + 2 = 10.
Choose yellow ball = 5/10.
Jaiprakash said:
9 years ago
@Obiano.
The answer is 1/2.
The answer is 1/2.
Tushar dhok said:
8 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)
Anitha said:
8 years ago
How it is possible?
(1)
Boopathy said:
8 years ago
Probability of selection Red or Green= 8+6=14 / 21 = 2/3.
Then how 1/3?
Then how 1/3?
(1)
Tharunteja said:
7 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.
Anitha said:
7 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.
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