To find the probability that three randomly selected students will have blood types A, B, and AB respectively, we multiply the individual probabilities of each blood type.
Given the probabilities:
- Probability of blood type A: \( P(A) = \frac{41}{100} \)
- Probability of blood type B: \( P(B) = \frac{1}{10} = \frac{10}{100} \)
- Probability of blood type AB: \( P(AB) = \frac{1}{25} = \frac{4}{100} \)
Now we will calculate the combined probability:
\[ P(A, B, AB) = P(A) \times P(B) \times P(AB) \]
Substituting the values:
\[ P(A, B, AB) = \left(\frac{41}{100}\right) \times \left(\frac{10}{100}\right) \times \left(\frac{4}{100}\right) \]
Now, let's do the multiplication step-by-step:
\[ = \frac{41 \times 10 \times 4}{100 \times 100 \times 100} \] \[ = \frac{1640}{1000000} \]
Now we can simplify this fraction:
\[ = \frac{164}{100000} \]
Thus, the probability that three randomly selected students will have blood types A, B, and AB respectively is:
\[ \frac{164}{100000} \]
Or, as a decimal:
\[ 0.00164 \]
Therefore, the probability is \( \frac{164}{100000} \) or \( 0.00164 \).