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Y_Kistochka [10]
3 years ago
8

Suppose that the inside bottom of a box is painted with three colors: 1/3 of the bottom area is red, 1/6 is blue, and 1/2 is whi

te. You toss a tiny pebble into the box without aiming and note the color on which the pebble lands. Then you toss another tiny pebble into the box without aiming and note the color on which that pebble lands. What is the probability that the first pebble lands on the color blue and the second lands on red? (Enter your probability as a fraction.)
Mathematics
1 answer:
Rus_ich [418]3 years ago
8 0

Answer:

The Probability of the pebble landing on red = 1/3

The Probability of the pebble landing on blue = 1/6

The Probability of the pebble landing on white = 1/2

Total sum of probability = 1/3 + 1/6 + 1/2 = 1

The probability that the first pebble lands on blue and second pebble lands on red is:

1/6 * 1/3

= 1/18    

Step-by-step explanation:

Red, Red = 1/3 * 1/3

Red, Blue = 1/3 * 1/6

Red, white = 1/3 * 1/2

Blue, Blue = 1/6 * 1/6

Blue, Red = 1/6 * 1/3

Blue, white = 1/6 * 1/2

White, White = 1/2 * 1/2

White, Red = 1/2 * 1/3

White, Blue = 1/2 * 1/6

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First you would do parenthesis according to the PEMDAS(Parenthesis, Exponents, Multiply, Divide, Add, Subtract) Rule

4 -2(-5a -10) = 30

4 + 10a + 20 = 30

Now you need to add 4 and 20 because they are on the same side

24 + 10a = 30

Now subtract 24 on both sides

24 + 10a = 30

-24. -24


10a = 6

Last divide 10 from both sides


10a/10. = 6/10

a= 6/10

Last simplify (optional but recommended)

a = 3/5 or 0.6


Hope this helped :)


A = 0.6
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Cohen's D

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