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Alexxandr [17]
3 years ago
15

A bag contains red balls and white balls. If five balls are to be pulled from the bag, with replacement, the probability of gett

ing exactly three red balls is 32 times the probability of getting exactly one red ball. What percent of the balls originally in the bag are red?
Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
6 0

Answer:

Percentage of balls which are red = 80%.

Step-by-step explanation:

Let the probability of drawing a red ball  be x then the probability of drawing a white ball is 1-x.

There are 5C3 = 10 ways of getting 3 reds in 5 draws so  the probability of this is 10* x^3 * (1 - x)^2.

The probability of getting 1 red in 5 draws = 5C1 * x (1-x)^4.

The  first probability is 32 times the last.

So we have the equation:

10x^3(1 - x)^2  /  5x(1 - x)^4  = 32

2x^2 / (1 - x)^2 = 32

2x^2 = 32(1 -x)^2

2x^2 = 32( 1 - 2x + x^2)

2x^2 = 32 - 64x + 32x^2

30x^2 - 64x  + 32 = 0

15x^2 - 32x + 16 = 0

(5x - 4)(3x - 4) = 0

x = 0.8, 1.333...

The probability must be < 1 so x = 0.8  = 80%.

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Answer:

B.166 11/12

Step-by-step explanation:

To handle this problem better, we can separate the fraction and integer portions to look at the whole situation better.

First, let's only look at the integers. Anthony flew up his drone 125 feet (translating to +125), allowed it to drop 14 feet (translating to -14), and then flew it up for 55 feet (translating to +55). Hence, only considering the integers, the total change in the drone's position would be (125 - 14 + 55) feet, which equates to 166 feet.

You'd probably have already noticed this answer is quite close to B. 166 11/12 feet, but let's keep on going for now.

On the fraction part, he flew his drone up 1/2 feet, allowed it to drop 1/3 feet, and then flew it up for 3/4 feet, translating to (1/2 - 1/3 + 3/4) feet.

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