In a bag there are 20 cards 10 names A and another 10 names B. Cards are drawn randomly one by one with replacement then find probability that second A comes before third B

Options

A. `15/16`
B. `11/16`
C. `9/16`
D. `13/16`

Answer

To solve the problem of finding the probability that the second A comes before the third B when drawing cards from a bag containing 10 cards labeled A and 10 cards labeled B, we can follow these steps:

### Step 1: Define the probabilities
The total number of cards is 20, with 10 cards labeled A and 10 cards labeled B. Therefore, the probability of drawing an A card (P(A)) and the probability of drawing a B card (P(B)) are both:

\[ P(A) = \frac{10}{20} = \frac{1}{2} \]
\[ P(B) = \frac{10}{20} = \frac{1}{2} \]

**Hint:** Remember that the probabilities of independent events can be calculated based on the total outcomes.

### Step 2: Understand the event
We need to find the probability that the second A appears before the third B. This means we are interested in sequences of draws where the second A occurs before the third B.

### Step 3: Analyze the sequences
To find the probability, we can consider the sequences of draws. We can denote the events as follows:
- A: A card is drawn
- B: B card is drawn

We want the sequences where the second A appears before the third B. The possible sequences can be represented as combinations of A's and B's.

### Step 4: Calculate the favorable sequences
Let’s denote the positions of the draws:
- The first two A's can occur in any of the first \( n \) draws, where \( n \) is the total number of draws until we reach the second A and the third B.

The sequences can be represented as:
- AAB (where the second A is before the third B)
- ABA
- BAA
- Any other combination that satisfies the condition.

### Step 5: Use combinatorial counting
The total number of ways to arrange the cards can be computed using combinations. The number of ways to arrange \( k \) A's and \( m \) B's is given by:

\[ \text{Total arrangements} = \frac{(k+m)!}{k!m!} \]

For our case, we need to find the arrangements that satisfy our condition.

### Step 6: Calculate the probability
The probability that the second A comes before the third B can be derived from the total arrangements of A's and B's. The total number of arrangements is \( 2^n \) (since each draw can either be A or B).

Using the symmetry of the problem, we can conclude that the probability of the second A appearing before the third B is given by:

\[ P(\text{second A before third B}) = \frac{11}{16} \]

### Final Answer
Thus, the probability that the second A comes before the third B is:

\[ \boxed{\frac{11}{16}} \]