Probability
Binomial Distribution — P(X≥7)
nta_pyq_2026_jan
Grade 12
Question:
From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is
67/10^8
7/10^7
81/10^8
73/10^8
Step-by-Step Solution
Key Concept: $n=8$, $p=\tfrac{1}{10}$, $q=\tfrac{9}{10}$. $P(X\geq7)=P(X=7)+P(X=8)$.
$P(X\geq7)=\dfrac{73}{10^8}$.
Correct Answer: 4