A bag has balls marked 0–9. Four balls drawn with replacement. Probability that **none** is marked '0' is:
- $(1/10)^4$
- $(9/10)^4$
- $4(9/10)^4$
- $1 - (9/10)^4$
Show answer and solution
Correct answer: B. $(9/10)^4$
$p = 1/10$ for '0'; $q = 9/10$ for 'not 0'.
$P(X = 0) = (9/10)^4 = 0.6561$.