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The theoretical probability of randomly picking each color marble is the number of color marbles you have for each color, divided by the total number of marbles. For example, the probability of selecting a red marble is 3/20.
3/6 or 1/2 or 50%
When you are asked what is the probability of drawing "thing a" when you have only the same amount of "thing b," is called an equally likely event. For example: In a hat you have 8 black marbles and 8 black marbles. Since there the same amount of both, the chance of picking a black marble is 50% and picking a blue marble is 50%. This is an equally likely event.
The probability of drawing a white marble is .46
There are 8 marbles that aren't black, out of a total of 12 marbles, so the probability is 8/12 or 2/3.
3/6 * 3/5 = 6/30 or 1/5 so you have a 20% chance of pulling a white and then black marble.
80% chance, Or 40/50
1 in 52
Number of possibilities for one category / Total of all possibilities. For example, if I had a bag of marbles where there are three white marbles and two black marbles. The probability of pulling out a white marble is how many white marbles are in the bag which is: three. But the total of things you can draw out of the bag can either be one of the three white marbles or one of the two black marbles. 3 white marbles+ 2 Black marbles= five marbles. Possibility is 3/5 for drawing a white marble.
a black one
It depends what probability exactly you want to find.probability = number of successful ways / total number of waysIf the problem is:You have a bag containing 4 blue, 5 red, 1 green, 2 black marble what is the probability of picking a blue marble at random?Thensuccessful ways = 4 as there are 4 blue marblestotal ways = 12 as there are 4 [blue] marbles + 5 [red] marbles + 1 [green] marble + 2 [black] marbles = 12 marbles in total.pr(picking a blue) = 4/12 = 1/3Perhaps the problem is:You pick 2 marbles at random without replacing them, what is the probability that they are the two black marbles?Each picking of a marble is an event and the two events are independent (in the sense that whatever you pick first does not affect the probability of the second pick) so you multiply the probability of each together:pr(1st black) = 2/12 = 1/6pr(2nd black) = 1/11 (there is 1 less black marble in the bag)pr(2 blacks) = 1/6 × 1/11 = 1/66Perhaps it is:You pick 2 marbles at random replacing the marble after the first pick, what is the probability of picking the same colour each time?This time there are 4 possible colours and the probabilities of 2 marbles the same is calculated for each (similar to above) and then they are added together to find the total probability of 2 marbles of the same colour:pr(blue) = 4/12 → pr(2 blue) = 4/12 × 4/12 = 16/144pr(red) = 5/12 → pr(2 red) = 5/12 × 5/12 = 25/144pr(green) = 1/12 → pr(2 green) = 1/12 × 1/12 = 1/144pr(black) = 2/12 → pr(2 black) = 2/12 × 2/12 = 4/144→ pr(2 the same colour) = pr(2 blue) + pr(2 red) + pr(2 green) + pr(2 black)= 16/144 + 25/144 + 1/144 + 4/144 = 46/144 = 23/72And so on.
Bag with 10 marbles: 3 orange, 5 black, 2 white.Rephrasing the question.If a marble is drawn from the bag, then returned to the bag, and a second marbleis drawn, what is the probability that the first marble turns out white and the secondmarble black ?The probability for a marble to come out white from the bag is:P(W) = 2/10 = 1/5The probability for a marble to come out black from the bag is:P(B) = 5/10 = 1/2The probability for a marble to come out white, put back in the bag and then take again a marble for a second time and turns out to be black is:P(B2|W1) = (1/5)∙(1/2) = 1/10 = 0.10 = 10 %