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A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)

A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)

Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\);

The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).

A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)

Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\);

The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).

Answer: C

Hey Bunuel,

Is the trick to simplifying that expression into 1/3 just simply seeing that the ratio of top to bottom is 1:3? Sometimes I get caught up doing a problem and try to simplify algebraically without "zooming out" for a second to see that. Would you mind showing how you would simplify that algebraically just to satisfy my curiosity?

A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)

Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\);

The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).

Answer: C

Hey Bunuel,

Is the trick to simplifying that expression into 1/3 just simply seeing that the ratio of top to bottom is 1:3? Sometimes I get caught up doing a problem and try to simplify algebraically without "zooming out" for a second to see that. Would you mind showing how you would simplify that algebraically just to satisfy my curiosity?

I doubt this is a 700 lvl question...Even I, not a lover or probability questions, solved it under 1 minute: B+2b+5 / 9B+15 3B+5 / 3(3B+5) => simplify by 3B+5 and get 1/3

Is it because only one ball is being selected that we are not applying the intersection part where both are possible??

Bunuel wrote:

Official Solution:

A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)

Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\);

The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).

Is it because only one ball is being selected that we are not applying the intersection part where both are possible??

Bunuel wrote:

Official Solution:

A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?

A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)

Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\);

The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).

Answer: C

Yes, we are selecting only one ball and we want it to be either blue or green.
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Yes indeed. I find it easy to forget about the basics when you see something visually confusing.

Again I appreciate that some people are really into math and left brained but put yourselves in the shoes of those who are right brained. It's like someone who speaks with someone they like and suddenly forget how to speak because they are nervous !

I had a doubt regarding the answer. We need to find the probability of getting a blue or a green ball. So 1st we could pick up green then blue or 1st we could pick up blue then green. so shouldn't 1/3 be multiplied by 2. If we were given that 1st green and second is blue then the answer 1/3 would be correct. Looking forward to hearing from you.

I had a doubt regarding the answer. We need to find the probability of getting a blue or a green ball. So 1st we could pick up green then blue or 1st we could pick up blue then green. so shouldn't 1/3 be multiplied by 2. If we were given that 1st green and second is blue then the answer 1/3 would be correct. Looking forward to hearing from you.

Regards

We are selecting only one ball and we want it to be either blue or green.
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