Can someone help me w/ “7 red balls, 6 blue balls”

[mathisnotforme]

Can someone help me w/ “7 red balls, 6 blue balls”

The question is telling me that I have 7 red balls and 6 blue balls from which randomly to select two without replacement and asking me what is the probability that the second ball is blue given that at least one of the balls is blue.


I know right off the bat that I'm working with Pr[second ball is blue|at least one of the balls is blue], but I am having trouble diagramming it. I know I'm working with P(A∩B)/P(B), but I am not sure what to calculate to get those variables. Can someone help me get started? Would the probability of the second being blue be 42/156+30/156?






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[Dr.Peterson]

The question is telling me that I have 7 red balls and 6 blue balls from which randomly to select two without replacement and asking me what is the probability that the second ball is blue given that at least one of the balls is blue.

I know right off the bat that I'm working with Pr[second ball is blue|at least one of the balls is blue], but I am having trouble diagramming it. I know I'm working with P(A∩B)/P(B), but I am not sure what to calculate to get those variables. Can someone help me get started? Would the probability of the second being blue be 42/156+30/156?
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Yes, that is P(second blue) = P(first red and second blue) + P(first blue and second blue).

You want P(second blue and at least one blue)/P(at least one blue), so your result is not directly needed (though ultimately it is), but you have most of the pieces.

What is P(at least one blue)?

What is P(second blue and at least one blue)?

An alternative method would be to use permutations: count the number of ways for the second ball to be blue and for at least one to be blue, and divide.

 

[pka]

The question is telling me that I have 7 red balls and 6 blue balls from which randomly to select two without replacement and asking me what is the probability that the second ball is blue given that at least one of the balls is blue.
I know right off the bat that I'm working with Pr[second ball is blue|at least one of the balls is blue], but I am having trouble diagramming it. I know I'm working with P(A∩B)/P(B), but I am not sure what to calculate to get those variables. Can someone help me get started? Would the probability of the second being blue be 42/156+30/156?​

I like to use subscripts to indicate order. \(\displaystyle \mathcal{P}(R_1R_2)=\dfrac{7\cdot 6}{13\cdot 12}=\dfrac{7}{26}\) is the probability that the first and the second balls are both red.
If \(\displaystyle \mathit{A}\) is the event that at least one of the two balls is blue then \(\displaystyle \mathcal{P}(\mathit{A})=1-\frac{7}{26}\). Can you explain this.
Can you finish?

\(\displaystyle \mathcal{P}(B_2|\mathit{A})=~?\)

 

[HallsofIvy]

There are 4 situations to consider: both balls are blue; first ball is blue, second
red; first ball is red, second blue; both balls are red.

Initially there are 6 blue balls and 7 red so the probability the first ball drawn is blue is 6/13. There are then 5 blue balls and 7 red so the probability the second ball drawn is also blue is 5/12. The probability the two balls are both blue is (6/13)(5/12)= 5/26.


Initially there are 6 blue balls and 7 red so the probability the first ball drawn is blue is 6/13. There are then 5 blue balls and 7 red so the probability the second ball drawn is red is 7/12. The probability the two balls are blue and red, in that order, is (6/13)(7/12)= 7/26.


Initially there are 6 blue balls and 7 red so the probability the first ball drawn is red is 7/13. There are then 6 blue balls and 6 red so the probability the second ball drawn is blue is 6/12= 1/2. The probability the two balls are red and blue, in that order, is (7/13)(1/2)= 7/26.

Initially there are 6 blue balls and 7 red so the probability the first ball drawn is red is 7/13. There are then 6 blue balls and 6 red so the probability the second ball drawn is also red is 6/12= 1/2. The probability both balls are red is (7/13)(1/2)= 7/26.

We are given that "at least one of the balls is blue" so we can ignore the last scenario. The total of the other three is 5/26+ 7/26+ 7/26= 19/26. The second ball is blue in the first and third scenarios. The probability that the second ball is blue, given that at least one of the balls is blue, is \(\displaystyle \frac{\frac{5}{26}+ \frac{7}{26}}{\frac{5}{26}+ \frac{7}{26}+ \frac{7}{26}}\)\(\displaystyle = \frac{\frac{12}{26}}{\frac{19}{26}}\)\(\displaystyle = \frac{12}{19}\).