12.
I drove my car over a bunch of nails. I estimate there’s a 10% chance that any given tyre
is punctured. I have only one spare tyre. The probability that I’ll be able to drive on
safely (after changing a tyre if necessary) is:
a. 95%
b. 90%
c. 66%
d. 60%
Probability of Driving Safely with One Spare Tire
The probability of driving safely with one spare tire is 95%, as this occurs when zero or exactly one of the four tires is punctured (10% puncture chance each, independent events).
Problem Breakdown
A standard car has four tires, each with a 10% (0.1) independent probability of puncture, so 90% (0.9) chance of being good. Driving safely requires changing at most one punctured tire using the spare, leaving three good tires plus the spare.
- The probability of zero punctures is 0.9^4 = 0.6561.
- The probability of exactly one puncture is binomial: C(4,1) × 0.1^1 × 0.9^3 = 4 × 0.1 × 0.729 = 0.2916.
- Total safe probability: 0.6561 + 0.2916 = 0.9477 or 95%.
Option Analysis
- a. 95%: Correct, matches the calculation for 0 or 1 puncture (safe scenarios).
- b. 90%: Incorrect; this might assume only one specific tire punctures or confuses with single-tire probability, ignoring multiples.
- c. 66%: Incorrect; approximates 0.94 (zero punctures only), forgetting the one-puncture case fixable by spare.
- d. 60%: Incorrect; no clear basis, possibly a miscalculation like 0.93.
Binomial Calculation Steps
- Probability all four tires good: 0.9^4 = 0.6561 or 65.61%.
- Probability exactly one punctured: 4 × 0.1 × 0.9^3 = 0.2916 or 29.16%.
- Total safe: 94.77%, rounds to 95% (option a).
Common Errors in Tire Puncture Math
- Overlooks spare: 66% assumes zero punctures only.
- Single tire focus: 90% ignores multiples.
- Real-world tire blowouts occur in ~11,000 U.S. crashes yearly, emphasizing probability awareness.
In probability car tires punctured spare tire scenarios, hitting nails creates a 10% puncture risk per tire across four tires, with one spare allowing safe driving if at most one fails. This classic binomial probability puzzle tests understanding of independent events and limits on fixes.
