Problem with Binomial distribution

Question

Can you help me with this problem - how to solve it or what methods to use?

1. The amount of burnt lamps in a device is a binomially distributed random variable $$X \sim \mathcal{Bin}(3, 1/3)$$ The probabilities of the device breaking after the failure of one, two or three lamps are respectively $$ p_{1} = 0.5, p_{2} = 0.6, p_{3} = 0.8$$
   • a) What is the probability of the device to break?
   • b) If the device is damaged, what is the probability that more than one lamp burned?

Answer 1

Let $A$ the event: the device breaks. You need to find $P(A)$.

Using the law of total probability, we can write: $$P(A)=\sum_{i=0}^3P(A|X=i)P(X=i)$$

If no lamp is burnt, then we can assume the probability for the device to break is $0$. So $P(A|X=0)=0$. For $i=1,2,3$, $P(A|X=i)$ is given to you, it's $p_i$.

Now you just have to find $P(X=i)$ for $1\leq i\leq 3$, using the fact that $X\sim B(3,1/3)$.