Respuesta :
For the first letter drawn, there are no conditions/restrictions we need to consider. So, the number of ways to draw an i is:
[tex]\text{Draw(i): } \frac{4}{11}[/tex]
For the second draw, the i is not replaced. That means, we've shrunk the sample space because we're dealing with less elements for the second draw. Firstly, one i disappears since we picked an "i" and there are 10 elements left, not 11.
Thus, to draw the second i, we have:
[tex]\text{Draw(ii): } \frac{3}{10}[/tex]
[tex]\text{P(drawing 2 i's): } \frac{4}{11} \cdot \frac{3}{10} = \frac{12}{110} = \frac{6}{55}[/tex]
[tex]\text{Draw(i): } \frac{4}{11}[/tex]
For the second draw, the i is not replaced. That means, we've shrunk the sample space because we're dealing with less elements for the second draw. Firstly, one i disappears since we picked an "i" and there are 10 elements left, not 11.
Thus, to draw the second i, we have:
[tex]\text{Draw(ii): } \frac{3}{10}[/tex]
[tex]\text{P(drawing 2 i's): } \frac{4}{11} \cdot \frac{3}{10} = \frac{12}{110} = \frac{6}{55}[/tex]
Answer:
For the first letter drawn, there are no conditions/restrictions we need to consider. So, the number of ways to draw an i is:
For the second draw, the i is not replaced. That means, we've shrunk the sample space because we're dealing with less elements for the second draw. Firstly, one i disappears since we picked an "i" and there are 10 elements left, not 11.
Step-by-step explanation:
