2019 Center Test probability: two bags, repeated draws, and conditioning

The original random experiment

This is a worked English explanation of Question 3 from the probability section of the 2019 Center Test Mathematics I/A. It is a historical Center Test problem, not a paper from the later Common Test.

A red bag contains two red balls and one white ball. A white bag contains one red ball and one white ball. Every draw is uniform among the balls in the selected bag, and the ball is returned after its color is observed.

For the first operation, roll a fair six-sided die. Choose the white bag if the result is a multiple of three, and the red bag otherwise. Draw and replace one ball.

For the second and third operations, choose the bag with the same color as the ball drawn in the immediately preceding operation, then draw and replace a ball again.

The questions ask for:

1. The first-operation probabilities of choosing each bag and obtaining a red ball.
2. The probability that the second operation uses the white bag.
3. The probabilities of a white ball on operations two and three, using the first white-ball probability $p$.
4. Two conditional probabilities after a white ball is observed.

Returning the ball means that each bag's composition stays fixed. The bag choice depends on the previous outcome, so the successive colors are not independent.

Part 1: the first red-ball probabilities

The die chooses the red bag with probability $4/6=2/3$, and a red draw from that bag has probability $2/3$. Thus

$$\boxed{P(\text{red bag and red ball on operation 1})=\frac23\cdot\frac23=\frac49.}$$

The white bag is selected with probability $2/6=1/3$, and contains a red ball with draw probability $1/2$. Hence

$$\boxed{P(\text{white bag and red ball on operation 1})=\frac13\cdot\frac12=\frac16.}$$

The total first red-ball probability is $4/9+1/6=11/18$.

Part 2: which bag is used next?

The second operation uses the white bag exactly when the first ball is white. Therefore

$$\boxed{P(\text{white bag on operation 2})=1-\frac{11}{18}=\frac7{18}.}$$

The probability here is the previous ball's color probability, not a fresh die-based probability of $1/3$.

Part 3: the recurrence-like calculation

Suppose the first white-ball probability is denoted by $p$. A second white ball can arise from two disjoint paths:

• First white, then a white draw from the white bag: probability $p/2$.
• First red, then a white draw from the red bag: probability $(1-p)/3$.

Their sum is

$$P(W_2)=\frac12p+\frac13(1-p)=\boxed{\frac16p+\frac13.}$$

Substitute $p=7/18$:

$$\boxed{P(W_2)=\frac16\cdot\frac7{18}+\frac13=\frac{43}{108}.}$$

The same transition applies between operations two and three:

$$\boxed{P(W_3)=\frac16\cdot\frac{43}{108}+\frac13=\frac{259}{648}.}$$

This is why the problem resembles a probability recurrence, even though only three operations are required.

Optional extension to more operations

If the rule continues and $p_n=P(W_n)$, then $p_{n+1}=p_n/6+1/3$. Its fixed point is $2/5$. Since $p_1=7/18$,

$$p_n=\frac25-\frac1{90}\left(\frac16\right)^{n-1}.$$

This reproduces all three probabilities and shows how the influence of the initial die choice diminishes.

Part 4a: infer the bag color from a white second ball

Question. Given that the second ball is white, what is the probability that it came from the white bag?

Let $W_i$ mean that operation $i$ produces a white ball. The second bag is white exactly when $W_1$ occurs, so the required probability is

$$P(W_1\mid W_2)=\frac{P(W_1\cap W_2)}{P(W_2)}.$$

The numerator is $(7/18)(1/2)$, and the denominator is $43/108$. Hence

$$\boxed{P(W_1\mid W_2)=\frac{(7/18)(1/2)}{43/108}=\frac{21}{43}.}$$

Part 4b: was the third ball the first white ball?

Question. Given that the third ball is white, what is the probability that it was the first white ball drawn?

The numerator requires the path red, red, white:

$$P(R_1\cap R_2\cap W_3)=\frac{11}{18}\cdot\frac23\cdot\frac13.$$

Dividing by the already calculated $P(W_3)$ gives

$$\boxed{P(R_1\cap R_2\mid W_3)=\frac{(11/18)(2/3)(1/3)}{259/648}=\frac{88}{259}.}$$

The numerator must specify the entire event “first white on operation three,” not merely the probability of a white third ball. For the general reasoning, see conditional probability.

Answers collected by part

• Part 1: $4/9$ and $1/6$.
• Part 2: $7/18$.
• Part 3: coefficient $1/6$, then $43/108$ and $259/648$.
• Part 4: $21/43$ and $88/259$.

The original test encodes these values in numbered answer boxes. The English explanation uses the values and subquestion numbers directly, so it can be read without the Japanese box labels.