Q.If the probability that a fluorescent light has a useful life of at least $600$ hours is $0.9$, find the probabilities that among $12$ such lights

Step 1. Identify the distribution. $n=12,\ p=0.9,\ q=0.1$; $X\sim B(12,0.9)$.

Step 2. (i) Exactly $10$. $P(X=10)=\dbinom{12}{10}(0.9)^{10}(0.1)^2=66\times(0.9)^{10}\times0.01$. Since $(0.9)^{10}\approx0.34868$: $P(X=10)\approx66\times0.34868\times0.01\approx0.2301$.

Step 3. (ii) At least $11$. Need $P(X=11)+P(X=12)$.

$P(X=11)=\dbinom{12}{11}(0.9)^{11}(0.1)=12\times(0.9)^{11}\times0.1$. $(0.9)^{11}=(0.9)^{10}\times0.9\approx0.31381$, so $P(X=11)\approx12\times0.31381\times0.1\approx0.37657$.

$P(X=12)=(0.9)^{12}=(0.9)^{11}\times0.9\approx0.28243$.

Sum: $P(X\ge11)\approx0.37657+0.28243=0.65900$. …

• Recomputing part (iii) from scratch instead of reusing $P(X=11),P(X=12)$ from part (ii) as $P(Y=1),P(Y=0)$ …