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3 red out of 9 marbles is a fraction of 3 out of 9 or 3/9 = 1/3 red

1/3 are red.

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6y ago
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6y ago

The fraction is 3/9.

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Q: What fraction of 9 marbles are red when there are 3 red?
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What is the number of each colour of marbles in a bag of 76 marbles if the ratio of red marbles to green marbles is 2 to 3 and the ratio of green marbles to blue marbles is 9 to 4?

The ratios are: red : green = 2 : 3 = (2×3) : (3×3) = 6 : 9 green : blue = 9 : 4 → ratio of red : green : blue = 6 : 9 : 4 There are 6 + 9 + 4 = 19 parts 76 marbles ÷ 19 parts = 4 marbles per part → red: 6 parts = 6 × 4 marbles per part = 24 red marbles → green: 9 parts = 9 × 4 marbles per part = 36 green marbles → blue: 4 parts = 4 × 4 marbles per part = 16 blue marbles To check: red + green + blue = 24 marbles + 36 marbles + 16 marbles = 76 marbles in the bag.


A marble is selected from a random jar containing 3 red marbles 4 yellow marbles and 2 marbles what are the odds in favor of selecting a red marble?

The probability of selecting a red marble is 3/9


A bag contains 5 red marbles 3 blue marbles and 1 green marble. A randomly drawn marble is not blue.?

it would be red because the probability is 5/9


There are red and blue marbles in a jar one-third of the marbles are red if another marble is added to the jar then 3 8 of the marbles will be red how many marbles are blue?

Depends on what the question is actually saying:If the "3 8" is 38 (thirty eight):Let R = number of red marbles to begin with.Let B = number of blue marblesGiven two facts:R = (1/3)(R+B)R+1 = 38. (This assumes it's a red marble added)2nd equation gives R = 371st equation gives B = 2R = 2*37 = 74 blueIf the "3 8" is 3 over 8 (3/8 or three eighths):If a non-red marble is added, the problem is unsolvable as 3/8 is greater than 1/3, but with more marbles and the same number of red marbles, the fraction of red marbles will be less than 1/3. Thus it is assumed that a red marble is added:Let M be the number of marbles in the jar, and R be the number of red ones, then:1/3 M = R -- one third are red3/8 (M + 1) = R + 1 -- when an extra red marble is added, 3/8 of the marbles are now red marblesSubstituting R from the first equation into the second equation gives and solving gives:3/8 (M + 1) = 1/3 M + 1→ 9/8 (M + 1) = M + 3→ 9(M + 1) = 8M + 24→ 9M + 9 = 8M + 24→ 9M = 8M + 15→ M = 151/3 are red → 2/3 are blue→ 2/3 x 15 = 10 are blue................................................................................................................................Another Answer:1/3+1/24 = 3/8 It doesn't how many red marbles are added because now there are 24 but before there were 23 which is 1/3 of 69 of which 2/3 are 46 blue marbles.


A bag contains 4 red marbles and 5 blue marbles If two marbles are chosen at random what is the probability of getting two red marbles?

1/6Because:There is a 4 in 9 chance of picking a red ball the first time and 3 in 8 chance of picking a red ball the second time. The chance of picking two reds as the first two balls is 4/9 x 3/8 or 12/72 or 1/6


Jan has 4 blues marbles 3 red marbles 2 yellow marbles in a bag Jan will randomly pick one marble from the bag What is the probability that the marble will be yellow?

2/9


A bag contains 4 red marbles and 3 black marbles You dram one marble look at it and put it back Then you draw one marble again What are your chances of getting 2 black marbles?

3/7*3/7= 9/49


What is the probability of pulling out a red marble then a blue marble without replacing the red marble the bag contains 3 red marbles 2 blue marbles and 5 green marbles?

1 out of 15 Probab. = Prob. of red x Prob. of blue Probab. = (3/10)x(2/9) = 5/90 = 1/15


What is the probability of choosing a red marble or a green marble out of a bag containing 5 blue marbles 3 red marbles and 1 green marble?

Pr = ways_of_success/total_ways Pr(Red) = 3/(5+3+1) = 3/9 Pr(Green) = 1/9 As they are independent: Pr(Red or Green) = Pr(Red) + Pr(Green) = 3/9 + 1/9 = 4/9 ---------------------------------------------------- Alternatively, as there are only three colours, selecting either Red or Green is the same as not selecting Blue: Pr(Red or Green) = Pr(not Blue) = 1 - Pr(Blue) → Pr(Red or Green) = 1 - 5/9 = 4/9


There are 9 red and 6 blue marbles in a jar If you draw red marbles from the jar 7 times what is the experimental probability of drawing a red marble?

The answer is dependent on whether of not you replace the marbles in the jar. If you do, the probability of drawing a red marble is 9 in 15 or 60%, every time. If you do not replace the marbles, the probability of drawing a red marble is 2 in 8 or 25%.


A bag contains 3 red marbles 2 blue marbles and 5 green marbles What is the probability of selecting a red marble without replacing it in the bag and then selecting a green marble?

Total marbles in the bag = 10Number of red ones = 3Probability of pulling a red one on the first draw = 3/10 = 0.3Total marbles remaining after the first draw = 9Number of green ones = 5Probability of pulling a green one after a red one has been withdrawn = 5/9Probability of both outcomes = (3/10) x (5/9) = (15/90) = 1/6 = (16 and 2/3) percent.


What is the ratio or rate as a fraction in simplest form for 9 goldfish out of 20 marbles?

9/20 is the simplest form of this fraction. Goldfish and Marbles aren't really comparable unless the goldfish is equivalent to a certain number of marbles. Say the Goldfish cost x and the marbles cost y, a goldfish is then x/y marbles so the answer is 9x/20y. As I don't know x or y, you'll have to simplify that yourself.