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podryga [215]
3 years ago
12

A game at the school carnival involves drawing colored marbles out of a bag for a prize. If you draw a red or blue marble, you w

in. If you draw a green or yellow marble, you lose. After each marble is drawn, it is put back in the bag.
You watch several people play with the following results:

G G G B Y G Y G G B G R B Y G R G B Y Y

Based on your observations, what is the probability that you will win?
A
0.20
B
0.30
C
0.45
D
0.70
Mathematics
1 answer:
mina [271]3 years ago
4 0
All up there are 20 marbles and only 6 of those are chances of winning so we make it into a fraction 6/20 then times 20 by five to get to 100 (what ever we do to the bottom we do to the top) times 6 by five giving you a fraction of 30/100 then divide 30 by 100 giving you 0.30 ANSWER EQUALS = 0.30
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Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$           $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

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$|\frac{1}{(n+1)!2^{n+1}}| \leq 10^{-4}$

$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$         (with error 0.0001)

 

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

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Step-by-step explanation:

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Step-by-step explanation:

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