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A bag contains six red , five white , and four
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24 Aug 2016, 18:51
24 Aug 2016, 18:51
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Question Stats:
70% (01:46) correct
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A bag contains six red , five white , and four blue balls and two balls are drawn at a random. The probability that none of them being red is
A. 1/35
B. 1/2
C. 12/35
D. None
Here this was my approach first I will calculate total number of balls which are 15
Then we have to choose 2 balls, So we can choose 15C2 so now we have to choose balls which should not include the red balls. so 5(white)+4(blue) = 9. so we can choose 9C2 ways.
Now 9C2/15C2=12/35.
It will be the ans.
But I am afraid that approach which you taught us,
I think I am taking in the wrong direction because If I choose
(None of them being red = 1 - All of them being red)
it is not working. I am getting the wrong ans. Please help me regarding this question Brent. How to solve this question by your approach.
A. 1/35
B. 1/2
C. 12/35
D. None
Here this was my approach first I will calculate total number of balls which are 15
Then we have to choose 2 balls, So we can choose 15C2 so now we have to choose balls which should not include the red balls. so 5(white)+4(blue) = 9. so we can choose 9C2 ways.
Now 9C2/15C2=12/35.
It will be the ans.
But I am afraid that approach which you taught us,
I think I am taking in the wrong direction because If I choose
(None of them being red = 1 - All of them being red)
it is not working. I am getting the wrong ans. Please help me regarding this question Brent. How to solve this question by your approach.
Re: A bag contains six red , five white , and four
[#permalink]
24 Aug 2016, 20:22
24 Aug 2016, 20:22
1
C is the answer.
The questions asking the probability of choosing a blue and white ball and not a red. So in the first pick the probability of choosing a blue or white is 9/15 as there are 6 red and 9 blue and white balls (totally no combined). The second pick will be 8/14 as one ball has been taken out in the first pick. Multiply the two and u have the answer. Hope this helps.
Cheers !!
The questions asking the probability of choosing a blue and white ball and not a red. So in the first pick the probability of choosing a blue or white is 9/15 as there are 6 red and 9 blue and white balls (totally no combined). The second pick will be 8/14 as one ball has been taken out in the first pick. Multiply the two and u have the answer. Hope this helps.
Cheers !!
Re: A bag contains six red , five white , and four
[#permalink]
24 Aug 2016, 21:53
24 Aug 2016, 21:53
I think this question is quite straightforward,here is my solution:
Total Balls in Bag = T = 6 red balls + 5 white balls + 4 blue balls = 15 balls
Now we are required to find the probability that neither of two balls that are drawn at random from bag is red.This comes out to be product of two probabilities.Probability of drawing first ball which is not red is independent of the probability of drawing second ball which is not red i.e.
P(1st ball drawn not red) * P(2nd ball drawn is not red) != 0 ... (a)
Both these events are independent and are not mutually exclusive
Total Number of Balls that are not red = 9
Probability of drawing two balls at random that are not red = P(1st ball drawn not red) * P(2nd ball drawn is not red)
= (9/15)*(8/14) = 12/35 ...(option C)
After drawing first ball (without replacement) which is not red we are left with 8 balls that are not red and in total we are left with 14 balls after drawing first ball.
Total Balls in Bag = T = 6 red balls + 5 white balls + 4 blue balls = 15 balls
Now we are required to find the probability that neither of two balls that are drawn at random from bag is red.This comes out to be product of two probabilities.Probability of drawing first ball which is not red is independent of the probability of drawing second ball which is not red i.e.
P(1st ball drawn not red) * P(2nd ball drawn is not red) != 0 ... (a)
Both these events are independent and are not mutually exclusive
Total Number of Balls that are not red = 9
Probability of drawing two balls at random that are not red = P(1st ball drawn not red) * P(2nd ball drawn is not red)
= (9/15)*(8/14) = 12/35 ...(option C)
After drawing first ball (without replacement) which is not red we are left with 8 balls that are not red and in total we are left with 14 balls after drawing first ball.
