92 of 100: Six Marbles

Suppose the marbles are labeled 1A, 1B, 2A, 2B, 3A, 3B. There are six possible choices of marble, and all equally likely.

$\begin{array}{cc|c} \hline \text{choice} & \text{color} & \text{other marble} \\ \hline 1A & red & red \\ 1B & red & red \\ 3A & red & blue \\ \hline 3B & blue & red \\ 2A & blue & blue \\ 2B & blue & blue \\ \hline \end{array}$

The observation that the chosen marble was red restricts the possibilities to the three at top of the table. Two of three have the "other marble" red; therefore the answer is $\boxed{2/3}$ .

Relevant wiki: Bayes' Theorem and Conditional Probability

We can do this quite easily using Bayes theorem:

$P(\text{Bag 1|Red})=\dfrac{P(\text{Red|Bag 1})\times P(\text{Bag1})}{P(\text{Red|Bag 1})\times P(\text{Bag 1})+P(\text{Red|Bag 3})\times P(\text{Bag 3})}=\dfrac{1\times \dfrac{1}{3}}{1 \times \dfrac{1}{3}+\dfrac{1}{2} \times\dfrac{1}{3}}=\boxed{\dfrac{2}{3}}$

Take R=red and B=blue. If the first pick is a red marble then possible outcomes are RR, RR and RB. In two of three of them second is also red. Thus the probability is $\frac{2}{3}$ .

2/3. There are six balls. All are equally likely to be the ball you picked. We're told you picked a red ball so we know you didn't take a ball from the bag with 2 blues and you didn't take the blue ball from the bag with one of each. That leaves 3 cases... One case is where the red ball you picked was in the bag with a blue ball. The other two cases are when you picked a red ball from the bag with two red balls. That means 2 out of 3 cases have another red ball in the bag.

This is a typical conditional probability problem.

The probability of drawing a red marble from bag 1 is 1/3 * (2/2) = 1/3

The probability of drawing a red marble from bag 2 is 1/3 * 0 = 0

The probability of drawing a red marble from bag 3 is 1/3 * (1/2) = 1/6

Therefore, using Bayes theorem, the probability that the marble came from the 1st bag (which is also the required probability) = 1/3 / (1/3 +1/6)

= 1/3 / 1/2

= 2/3

P(bag 1| we picked a red ball) = $\frac{P(bag1 and red)}{P(we picked a red ball from bags 1 or 3)}$ . P(bag1 and red) =P(red}bag 1)P(bag 1) = 1 $\times\frac{1}{3}$ = $\frac{1}{3}$ . But P(1st red) = P(red from bags 1, or 3) = P(red | bag 1)P(bag 1) + P(red | bag 3)P(bag 3) = $1\times \frac{1}{3} + \frac{1}{2} \times \frac {1}{3}$ = $\frac{1}{2}$ . Therefore, P(bag 1| red) = $\frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$ .

P(B given A)=P(B and A)/P(A).

The probability that you picked a red marble is (1+0.5+0)/3, which is 1/2.

If you choose the bag with two red marbles, you will always pick a red marble, so the probability that you picked a red marble out of the bag with two red marbles is the same as the probability that you chose the bag with two red marbles, which is 1/3.

Now, use the formula provided at the beginning: (1/3)/(1/2)=2/3

bags with red marbles is=2

total bags=3

so, possibility of getting red marbles is= $\frac{2}{3}$

This is not an elegant way to solve it, but it is the most intuitive.

In two bags we have two ball with same color, but they are different balls, so let's assume that the balls are numbered.

We have the following possibilities:

$\text{Marble left in the bag} \rightarrow \text{Drawn marble}$

Bag #1

$\color{#D61F06} \text{red marble \#1} \color{#333333} \rightarrow \color{#D61F06} \text{red marble \#2}$

$\color{#D61F06} \text{red marble \#2} \color{#333333} \rightarrow \color{#D61F06} \text{red marble \#1}$

Bag #2

$\color{#3D99F6} \text{blue marble \#1} \color{#333333} \rightarrow \color{#3D99F6} \text{blue marble \#3}$

$\color{#3D99F6} \text{blue marble \#2} \color{#333333} \rightarrow \color{#3D99F6} \text{blue marble \#1}$

Bag #3

$\color{#D61F06} \text{red marble} \color{#333333} \rightarrow \color{#3D99F6} \text{blue marble}$

$\color{#3D99F6} \text{blue marble} \color{#333333} \rightarrow \color{#D61F06} \text{red marble}$

The problem says that a red marble was drawn. Then, we can remove the cases where a blue marble is drawn. We have these cases left:

$\color{#D61F06} \text{red marble \#1} \color{#333333} \rightarrow \color{#D61F06} \text{red marble \#2}$

$\color{#D61F06} \text{red marble \#2} \color{#333333} \rightarrow \color{#D61F06} \text{red marble \#1}$

$\color{#3D99F6} \text{blue marble} \color{#333333} \rightarrow \color{#D61F06} \text{red marble}$

The problem asks for the probability of the other ball be red. Then, we have 2 cases in 3 where the ball left in the bag is red. Then the answer is $\boxed{\frac{2}{3}}$

There are 3 ways you can pick a red marble: $2$ ways to get a red marble from the first bag (there are $2$ red marbles in the first bag, and you can get any of them) and $1$ way to get it from the third bag. The only time the second marble would also be red is if it were from the first bag, so we get $\boxed{\frac{2}{3}}$ (Because the probability of something happening is equal to the number of cases of it happening over the total number of cases).

Label the three red marbles R1 R2 and R3. If you select a red marble at random (the initial problem setup gets you to this point), what is the probability that it's next to another red marble? 2/3 :-)

Since we know at least one of the marbles in the bag is red, it means that the bag picked was not Bag 2(which has 2 blue and 0 red marbles). Of the remaining 4 marbles in bags 1 and 3, 3 are red. If we drew either of the red marbles from Bag 1, the other is also red, but if we drew the one red marble from Bag 3, the other marble is blue, meaning that if we draw either of the red marbles from Bag 1 from the 3 red marbles, we have a successful outcome(the other is also red). Therefore, the probability is $\frac{2}{3}$ .

$P(\text{Red 2 | Red 1}) = \frac{P \text{( Red 2} \space \cap \space \text{Red 1)}}{P \text{( Red 1)}} = \frac{P \text{(Bag 1)}}{\frac{3}{6}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$