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Two marbles are drawn, without replacement, from an urn containing 4 red marbles, 5 white marbles, and 2 blue marbles. Determine the probability that:

a. Both are red.
b. The first marble drawn is red, and the second is blue.
c. The first marble drawn is blue, and the second is red.
d. One of the marbles is red and the other is blue.

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Thomas L. | Mathematics TutorMathematics Tutor
4.9 4.9 (26 lesson ratings) (26)
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Probability is the # of possible outcomes desired /  # of possible total outcomes.  In problems like this when you have consecutive draws, you find the probability of each individual draw and then multiply your results.
a. there are 4 reds to possible draw from the urn, there are 11 total marbles.  The probability of drawing 1 red is 4/11. 
After the 4/11 chance that you did get a red, now what is the probability of getting a second red? 
*Note, the was no replacing so you have 1 red in your hand and 3 in the urn. 
Now there is a 3/10 probability of getting the second red marble. 10 because you have one of the 11 in your hand so there are only 10 left in the urn. 
so we have 4/11 and 3/10, multiply and we get 12/110 which reduces to 6 /55
b. same idea just keep your colors and number of marbles straight. 
P(red) = 4/11
P(b after a red has been drawn) = 2/10, 2 blue still in the urn but only 10 marbles since you have 1 red in your hand.
Multiply 4/11 and 2/10 which equals 8/110 = 4/55
c. P(b) = 2/11, P(red after a blue has been drawn) = 4/10,  multiply 8/110 = 4/55
d. this one is different, you have to account for either one being red or blue.  so a red then blur or a blue then a red.  We have already found these probabilities in b. and c.  Since both situations need to be counted then we add them together.  4/55 +4/55 = 8/55
Why don't we multiply?  Because that would give us a smaller probability and if you think about the problem, out chances increase if the order doesn't matter. (yes there is a mathematical explanation but I like  to use common sense when possible)
Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
a) (4/11)(3/10) = 6/55
b) (4/11)(2/10) = 4/55
c) (2/11)(4/10) = 4/55
d) 2(4/55) = 8/55
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
Remember: P(A and B)=P(A)P(B) for independent events (which marble drawings are)
P(A or B)=P(A)+P(B) for disjoint events
1. P(red and red) =(4/11)(3/10)=12/110
2. P(red and blue)=(4/11)(2/10)=8/110
3. P(blue and red)=(2/11)(4/10)=8/110
4. P(red or blue)=P(red and blue)+P(blue and red)=16/110