A bag contains `4` tickets numbered `1,` `2,` `3,` `4` and another bag contains `6` tickets numbered `2,` `4,` `6,` `7,` `8,` `9`. One bag is chosen and a ticket is drawn. The probability that the ticket bears the number `4`, is equal to

Options

A. `1//48`
B. `1//8`
C. `5//24`
D. none of these

Answer

To solve the problem step by step, we will calculate the probability of drawing a ticket numbered 4 from either of the two bags.

### Step 1: Identify the bags and their contents
- Bag A contains tickets numbered: 1, 2, 3, 4 (4 tickets)
- Bag B contains tickets numbered: 2, 4, 6, 7, 8, 9 (6 tickets)

### Step 2: Calculate the probability of choosing each bag
Since there are two bags, the probability of choosing either bag is:
- Probability of choosing Bag A = \( \frac{1}{2} \)
- Probability of choosing Bag B = \( \frac{1}{2} \)

### Step 3: Calculate the probability of drawing a ticket numbered 4 from each bag
- For Bag A:
- There is 1 ticket numbered 4 out of 4 total tickets.
- Probability of drawing a 4 from Bag A = \( \frac{1}{4} \)

- For Bag B:
- There is 1 ticket numbered 4 out of 6 total tickets.
- Probability of drawing a 4 from Bag B = \( \frac{1}{6} \)

### Step 4: Use the law of total probability
The total probability of drawing a ticket numbered 4 can be calculated by considering both cases (choosing Bag A or Bag B):

\[ P(\text{4}) = P(\text{A}) \times P(\text{4 | A}) + P(\text{B}) \times P(\text{4 | B}) \]

Substituting the values we calculated:

\[ P(\text{4}) = \left(\frac{1}{2} \times \frac{1}{4}\right) + \left(\frac{1}{2} \times \frac{1}{6}\right) \]

### Step 5: Simplify the expression
Calculating each term:

1. From Bag A:
\[ P(\text{A}) \times P(\text{4 | A}) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \]

2. From Bag B:
\[ P(\text{B}) \times P(\text{4 | B}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \]

Now, we need to add these two probabilities:

\[ P(\text{4}) = \frac{1}{8} + \frac{1}{12} \]

### Step 6: Find a common denominator and add the fractions
The least common multiple of 8 and 12 is 24. We convert each fraction:

- Convert \( \frac{1}{8} \) to have a denominator of 24:
\[ \frac{1}{8} = \frac{3}{24} \]

- Convert \( \frac{1}{12} \) to have a denominator of 24:
\[ \frac{1}{12} = \frac{2}{24} \]

Now we can add them:

\[ P(\text{4}) = \frac{3}{24} + \frac{2}{24} = \frac{5}{24} \]

### Final Answer
The probability that the ticket bears the number 4 is \( \frac{5}{24} \).

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