Understanding the Probability Problem

The question asks for the probability of a specific event occurring on the second draw from a set of items, where the first item drawn is not replaced. We have a box containing ten chits, numbered from 0 to 9. This means there are 10 distinct chits initially.

Analyzing the Drawing Process

The process involves two steps:

1. A first chit is drawn at random from the ten chits and kept aside. This means there are now 9 chits remaining in the box.
2. A second chit is drawn at random from the remaining 9 chits.

We want to find the probability that the second chit drawn is the chit numbered “9”.

Calculating the Probability of the Second Chit Being "9"

Let S be the set of the initial chits: S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The total number of chits is 10.

We can approach this problem by considering the possible outcomes for the first draw and how they affect the second draw. Let A be the event that the first chit drawn is '9', and A' be the event that the first chit drawn is not '9'. Let B be the event that the second chit drawn is '9'. We want to find P(B).

We can use the law of total probability:

\( P(B) = P(B|A)P(A) + P(B|A')P(A') \)

Let's calculate the probabilities needed:

• P(A): Probability the first chit is '9'.
  There is 1 chit '9' out of 10 total chits.
  \( P(A) = \frac{1}{10} \)
• P(A'): Probability the first chit is not '9'.
  There are 9 chits that are not '9' out of 10 total chits.
  \( P(A') = \frac{9}{10} \)
• P(B|A): Probability the second chit is '9' given the first chit was '9'.
  If the first chit was '9', there are 9 chits remaining, none of which is '9'.
  \( P(B|A) = \frac{0}{9} = 0 \)
• P(B|A'): Probability the second chit is '9' given the first chit was not '9'.
  If the first chit was not '9', there are 9 chits remaining, one of which is '9'.
  \( P(B|A') = \frac{1}{9} \)

Now, substitute these probabilities into the formula for P(B):

\( P(B) = P(B|A)P(A) + P(B|A')P(A') \)

\( P(B) = (0) \times \left(\frac{1}{10}\right) + \left(\frac{1}{9}\right) \times \left(\frac{9}{10}\right) \)

\( P(B) = 0 + \frac{9}{90} \)

\( P(B) = \frac{1}{10} \)

Alternatively, by symmetry, in any sequence of draws without replacement, any specific chit is equally likely to be in any position (first, second, third, etc.). Since there are 10 equally likely chits initially, the probability that the chit '9' is in the second position is the same as the probability it is in the first position, which is \( \frac{1}{10} \).

Conclusion

The probability that the second chit drawn is “9” is \( \frac{1}{10} \).

This matches one of the given options.

Probability Concepts Revision

Let's summarize the key concepts used in this probability problem.

Additional Information on Probability Problems

Probability problems often involve understanding whether events are independent or dependent, and whether drawing is done with or without replacement. In this case, the draws are dependent events because drawing the first chit changes the set of chits available for the second draw. This is a classic example of sampling without replacement.

Understanding how the sample space (the set of possible outcomes) changes after each step is crucial for solving problems involving drawing without replacement.

For the first draw, there were 10 possible outcomes (chits 0 through 9). For the second draw, there were only 9 possible outcomes, and the specific outcomes depended on what was drawn first.

The method using the law of total probability breaks down the problem into scenarios based on the first draw, making it explicit how the outcome of the first draw impacts the probability of the second draw.