Calculating Conditional Probability of Exponential Lifetime

Question

The lifetime of a cell phone battery T is exponentially distributed with mean of 3 days. What is the probability that the battery survives more than 2 days given that it survived 1 day?

Attempt:

1) E(X) = 3 = $\frac{1}{\lambda}$; then $\lambda = \frac{1}{3}.$

2) Since this question is asking for $P(X\ge 2 | X=1)$, use the Conditional Probability equation.

3) Then this becomes $\frac{P( X \ge 2)}{P(X=1)}$, this simplifies to $\frac {e^-\lambda x}{1-e^{-\lambda x}}$ which gives about 1.811.

However, the answer is 0.7

Why?

Answer 1

In (2) the question is asking for $P(X \ge 2 | X \ge 1).$ Notice that $P(X \ge 2 | X = 1) = 0;$ failure at time 1 precludes the possibility of survival beyond time 2.

The no-memory property of the exponential distribution says that $P(X \ge 2 | X \ge 1) = P(X \ge 1).$ I think you can get the desired answer from there.

If you are familiar with R statistical software, the following demonstration with a million such exponential experiments may be helpful:

 lambda = 1/3;  m = 10^6
 x = rexp(m, lambda)
 cond = (x >= 1)  # logical 
 > mean(x[cond] >= 2)
 [1] 0.717887     # approximates  0.7165313