What is the probability of drawing a red and a blue marble?

Question

For the experiment of drawing 2 marbles without replacement from a jar containing two red marbles, 3 blue marbles, 4 green marbles and 1 mauve marble.. find each of the following probabilities? .....This red and a blue is throwing me off. I don't know how to set this up to come up with an answer.

Answer 1

There are ten marbles in the jar and thus there are 10!/((8!) (2!))=45 different pairs of marbles. Since there are 2 red marbles and 3 blue ones, then (2)(3) = 6 of these pairs are the red-blue combination. So the probability is 6/45 = 2/15.

In the other answer above, you need to add the two probabilities of the two cases together; thus 1/15 + 1/15 = 2/15

Answer 2

Find the total number of marbles. 2 red, 3 blue, 4 green, and 1 mauve.

That's 10 marbles.

So what's the probability of drawing the first red marble? Its 2:10 right? or 1:5

Now we have 9 marbles left: 1 red, 3 blue, 4 green and 1 mauve.

What's the probability of drawing the second blue marble? Its 3/9 or 1:3

So if you mulitiply the two: 1:5 x 1:3 then you're odds of drawing a red marble and then a blue one are 1:15

It should work the opposite way too... blue first than red:

3:10 x 2:9 = 6:90 = 1:15

Answer 3

For the 1st step, you should draw a crimson marble. There are 13 crimson marbles and an entire of three+13+a million=17 marbles so the prospect of having the crimson marble is 13/17. then you definately replace the marble lower back so we nevertheless have 17 marbles. yet this time you should pull the only a million yellow marble. So the prospect of pulling out the yellow marble is a million/17. on the grounds which you prefer first to pull the crimson marble and then the yellow marble, you multiply the possibilities, i.e., (13/17) *(a million/17)= 13/289 So the respond is A) Note1: I emphasize the be conscious and because it is the biggest be conscious that tells me to multiply the two possibilities. If the be conscious became OR extra complicated math will ought to be completed. be conscious 2: The replace is substantial by way of fact in case you probably did not replace the marble you will not have 17 marbles for the 2d danger. be conscious 3: If it is for a SAT form try, you should go with A by suggestions-set as follows: in the two circumstances you would be putting off marbles out of 17. So the denominator could be 17*17 (2 drawings). Now 17 is a prime selection so the numerator will purely cancel out multiples of 17. So the span of possible solutions could have a denominator of 17*17 or its multiples that are 17 and a million. so C and D are out at first look. To do away with B in simple terms calls for you to pull out a calculator and be conscious that 153/17=9 and 9 does not divide 28, so it may't be canceled out. That leaves answer A) it is not an much less annoying thank you to respond to the question -- yet while it is a attempt to additionally you're pressed for time, this methodology facilitates you to do away with C and D fullyyt and makes for a extra perfect probabilities of guessing properly on the grounds that your options are constrained to between a and b (2 options i.e. a million/2 danger to be desirable vs 4 options with a million/4 danger to be desirable)

Answer 4

red & blue = 2/10*3/9 = 6/90
blue & red = 3/10*2/9 = 6/90

12/90 = 2/15