## Probability of seeing a car in 10 minutes

### Solution

Let’s consider every \(10\) minutes interval like a coin flip. Let \(p\) the probability of not seeing a car in \(10\) minutes.

Let’s find the probability of not seeing a car in \(20\) minutes. It’s like repeating twice the event not seeing a car in \(10\) minutes.

\(Pb\)(no cars in \(20\)) = \(Pb\)(no cars in \(10\)) \(Pb\)(no cars in \(10\))

\(Pb\)(no cars in \(20\)) = \(p\)(\(p\))

\(Pb\)(no cars in \(20\)) = \(p^{2}\)

Now, let’s find the probability of the event not seeing a car in \(30\) minutes. It’s the probability of the event not seeing a car in \(20\) minutes, followed by the event not seeing a car in \(10\) minutes.

\(Pb\)(no cars in \(30\)) = \(Pb\)(no cars in \(20\)) \(Pb\)(no cars in \(10\))

\(Pb\)(no cars in \(30\)) = \(p^{2}\)(\(p\))

\(Pb\)(no cars in \(30\)) = \(p^{3}\)

we can directly calculate

\(Pb\)(no cars in \(30\)) = \(Pb\)(no cars in \(10\)) \(Pb\)(no cars in \(10\)) \(Pb\)(no cars in \(10\))= \(p^{3}\)

Let’s continue, the probability of seeing at least \(1\) car in \(30\) minutes is the complement event of seeing no cars in \(30\) minutes.

\(Pb\)(car in \(30\)) = \(1\) – \(Pb\)(no cars in \(30\))

\(Pb\)(car in \(30\)) = \(1\) – \(p^{3}\)

This is equal to \(0.95\). We get an equation to solve for \(p\) :

\(1\) – \(p^{3}\) = \(0.95\)

\(p^{3}\) = \(0.05\)

\(p\) = \(0.05^{\frac{1}{3}}

\) \(\approx

\) \(0.368

\)

The probability of seeing at least \(1\) car in \(10\) minutes is the complement event of seeing no cars in \(10\) minutes.

\(Pb\)(car in \(10\)) = \(1\) – \(Pb\)(no cars in \(10\))

\(Pb\)(car in \(10\)) = \(1\) – \(p\)

\(Pb\)(car in \(10\)) \(\approx

\) \(0.632

\)

Finally, the answer is approximately \(63\) percent.