Answer :
To find the probability that both beads drawn from the bag are white, let's follow these steps carefully:
1. Total Number of Beads: The bag contains \( n \) beads in total.
2. Number of White Beads: Given that there is one black bead, the number of white beads in the bag is \( n - 1 \).
3. First Bead is White:
- The probability of drawing a white bead first is calculated by considering the number of favorable outcomes (white beads) over the total number of outcomes (total beads in the bag).
- Therefore, the probability of drawing a white bead first is:
[tex]\[ \frac{n - 1}{n} \][/tex]
4. Second Bead is White:
- Once the first white bead has been drawn, there are now \( n - 1 - 1 = n - 2 \) white beads left in the bag.
- The total number of beads left in the bag is now \( n - 1 \) (one bead has already been taken out).
- Consequently, the probability of drawing another white bead is:
[tex]\[ \frac{n - 2}{n - 1} \][/tex]
5. Both Beads are White:
- Since the drawing of beads is without replacement, both events are dependent. The total probability that both beads drawn are white is the product of the probability of the first event and the probability of the second event:
[tex]\[ \text{Probability of two white beads} = \left( \frac{n - 1}{n} \right) \times \left( \frac{n - 2}{n - 1} \right) \][/tex]
6. Simplifying the Expression:
- Notice that the \( n - 1 \) terms cancel out in the fraction:
[tex]\[ \left( \frac{n - 1}{n} \right) \times \left( \frac{n - 2}{n - 1} \right) = \frac{(n - 1) \cdot (n - 2)}{n \cdot (n - 1)} = \frac{n - 2}{n} \][/tex]
Therefore, the probability that both beads drawn from the bag are white is:
[tex]\[ \boxed{\frac{n-2}{n}} \][/tex]
1. Total Number of Beads: The bag contains \( n \) beads in total.
2. Number of White Beads: Given that there is one black bead, the number of white beads in the bag is \( n - 1 \).
3. First Bead is White:
- The probability of drawing a white bead first is calculated by considering the number of favorable outcomes (white beads) over the total number of outcomes (total beads in the bag).
- Therefore, the probability of drawing a white bead first is:
[tex]\[ \frac{n - 1}{n} \][/tex]
4. Second Bead is White:
- Once the first white bead has been drawn, there are now \( n - 1 - 1 = n - 2 \) white beads left in the bag.
- The total number of beads left in the bag is now \( n - 1 \) (one bead has already been taken out).
- Consequently, the probability of drawing another white bead is:
[tex]\[ \frac{n - 2}{n - 1} \][/tex]
5. Both Beads are White:
- Since the drawing of beads is without replacement, both events are dependent. The total probability that both beads drawn are white is the product of the probability of the first event and the probability of the second event:
[tex]\[ \text{Probability of two white beads} = \left( \frac{n - 1}{n} \right) \times \left( \frac{n - 2}{n - 1} \right) \][/tex]
6. Simplifying the Expression:
- Notice that the \( n - 1 \) terms cancel out in the fraction:
[tex]\[ \left( \frac{n - 1}{n} \right) \times \left( \frac{n - 2}{n - 1} \right) = \frac{(n - 1) \cdot (n - 2)}{n \cdot (n - 1)} = \frac{n - 2}{n} \][/tex]
Therefore, the probability that both beads drawn from the bag are white is:
[tex]\[ \boxed{\frac{n-2}{n}} \][/tex]
