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nasty-shy [4]
3 years ago
12

A bag of uninflated balloons contains 10 red, 12 blue, 15 yellow and 8 green balloons. A balloon is drawn at random, what is the

probability of drawing a red balloon?
Mathematics
2 answers:
disa [49]3 years ago
5 0
Add all together ;10+12+15+8=35 probability is 10/35(.2857) 
(28.57%)
tia_tia [17]3 years ago
3 0

We have been given that a bag of uninflated balloons contains 10 red, 12 blue, 15 yellow and 8 green balloons.

We need to find the probability of drawing a red balloon.

We know that probability is given by

\\
\text{Probability =} \frac{\text{Favorable outcomes}}{\text{Total outcomes}}

Since we know that there are 10 red balloons. Therefore, our number of favorable outcomes is 10.

In order to find the total number of outcomes, we will add balloons of all colors.

Therefore, total outcomes are  = 10+12+15+8 = 45

We can now substitute the values of favorable outcomes and total outcomes in order to find the required probability.

\\
\text{Probability =} \frac{\text{10}}{\text{45}} = \frac{2}{9}

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defon
I'm guessing the series is supposed to be

\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}

By the ratio test, the series converges if the following limit is less than 1.

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|

The first n terms in the numerator's denominator cancel with the denominator's denominator:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|

|x^n| also cancels out and the remaining factor of |x| can be pulled out of the limit (as it doesn't depend on n).

\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0

which means the series converges everywhere (independently of x), and so the radius of convergence is infinite.
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Step-by-step explanation:

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