## Probability of seeing a car in 10 minutes

Home -> Solved problems -> 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.

Home -> Solved problems -> Probability of seeing a car in 10 minutes

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