Aptitude  Probability  Discussion
Discussion Forum : Probability  General Questions (Q.No. 2)
2.
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Answer: Option
Explanation:
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S)  = Number of ways of drawing 2 balls out of 7  
= ^{7}C_{2} `  


= 21. 
Let E = Event of drawing 2 balls, none of which is blue.
n(E)  = Number of ways of drawing 2 balls out of (2 + 3) balls.  
= ^{5}C_{2}  


= 10. 
P(E) =  n(E)  =  10  . 
n(S)  21 
Discussion:
112 comments Page 1 of 12.
Sneha said:
3 months ago
Can anyone explain this in the short method?
(4)
Riafat Hussain said:
4 months ago
1st ball = 5/7.
2nd ball = 4/6.
5/7 * 4/6 = 10/21.
2nd ball = 4/6.
5/7 * 4/6 = 10/21.
(21)
Ahmed Raafat said:
10 months ago
7 CHOOSE 2 can also be denoted as 7C2.
7 is the total number of distinct elements (n),
2 is the number of elements drawn or chosen at a time (k),
21 is the total number of possible combinations (C).
nCk = n! / (k! * (n  k)!)
7C2 =7! / (2! x 5!) = 21.
7 is the total number of distinct elements (n),
2 is the number of elements drawn or chosen at a time (k),
21 is the total number of possible combinations (C).
nCk = n! / (k! * (n  k)!)
7C2 =7! / (2! x 5!) = 21.
(5)
SA said:
11 months ago
Choose the first ball: There are 7 balls in total (2 red, 3 green, and 2 blue) , so you have 7 choices for the first ball.
Choose the second ball: After picking the first ball, there are 6 balls left for the second ball (since you're not concerned with color at this point).
Now, multiply the number of choices for the first and second balls together to get the total number of ways to draw any two balls: 7 choices × 6 choices = 42 ways.
Finally, calculate the probability that none of the balls drawn is blue by dividing the number of ways to draw without blue balls by the total number of ways to draw any two balls:
Probability = (Number of ways to draw without blue balls) / (Total number of ways to draw any two balls).
Probability = 20 / 42.
You can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:
Probability = (20 ÷ 2) / (42 ÷ 2).
Probability = 10 / 21.
So, the probability that none of the balls drawn is blue is 10/21.
Choose the second ball: After picking the first ball, there are 6 balls left for the second ball (since you're not concerned with color at this point).
Now, multiply the number of choices for the first and second balls together to get the total number of ways to draw any two balls: 7 choices × 6 choices = 42 ways.
Finally, calculate the probability that none of the balls drawn is blue by dividing the number of ways to draw without blue balls by the total number of ways to draw any two balls:
Probability = (Number of ways to draw without blue balls) / (Total number of ways to draw any two balls).
Probability = 20 / 42.
You can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:
Probability = (20 ÷ 2) / (42 ÷ 2).
Probability = 10 / 21.
So, the probability that none of the balls drawn is blue is 10/21.
(15)
Anhad said:
12 months ago
@Saiteja
I also thought that 20/21 is correct, actually, you have only deducted the probability of both blue from 1, you have to also deduct at least one blue from 1 i.e. 10/21. So the final calculation will go 1  1/21  10/21 = 10/21. (Probability of not getting blue = 1  the probability of both blue  probability of one blue)
I also thought that 20/21 is correct, actually, you have only deducted the probability of both blue from 1, you have to also deduct at least one blue from 1 i.e. 10/21. So the final calculation will go 1  1/21  10/21 = 10/21. (Probability of not getting blue = 1  the probability of both blue  probability of one blue)
(10)
Saiteja said:
12 months ago
A bag contains 2 red, 3 green and 2 blue balls, total = 7 balls.
(probability of choosing+probability of not choosing=1).
Let's consider the probability of choosing two balls is blue then,
probability of choosing P=(2/7)*(1/6) = 1/21.
so, the probability of not choosing (NP) = 1P,
NP = 1(1/21)
NP = 20/21,
I think 20/21 is the right answer.
(probability of choosing+probability of not choosing=1).
Let's consider the probability of choosing two balls is blue then,
probability of choosing P=(2/7)*(1/6) = 1/21.
so, the probability of not choosing (NP) = 1P,
NP = 1(1/21)
NP = 20/21,
I think 20/21 is the right answer.
(10)
Ronak Mistry said:
1 year ago
Total Balls = 7.
Num of Selected Balls are 2.
So, the Total Outcomes = 7C2.
Num of Blue Balls are Selected = 5C2.
So, the Probability = 5C2/7C2 = 10/21.
Num of Selected Balls are 2.
So, the Total Outcomes = 7C2.
Num of Blue Balls are Selected = 5C2.
So, the Probability = 5C2/7C2 = 10/21.
(11)
Tejeshwari said:
2 years ago
5/7 is it right?
Anyone can explain it, please.
Anyone can explain it, please.
(8)
Lyi said:
4 years ago
The probability of the first ball not being blue is 5/7 and the probability of the second ball also not being blue is 4/6 and this is so because we have already drawn one ball and now the total number of balls got reduced to 6. Now where the 4 comes from? Its because we're assuming that the first ball we've drawn is a nonblue and therefore we have one less ball than 5 to draw from now.
I hope it was helpful.
I hope it was helpful.
(22)
SRZ said:
4 years ago
P1st (not blue)= 1P(blue) = 12/7 = 5/7.
P2nd(not blue)= 1P(blue) = 12/6 = 4/6.
P1 and P2(not blue)=5/7 x 4/6 = 20/42 = 10/21.
P2nd(not blue)= 1P(blue) = 12/6 = 4/6.
P1 and P2(not blue)=5/7 x 4/6 = 20/42 = 10/21.
(44)
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