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