Question

Suppose four students miss an exam. They tell their instructor that they were carpooling to school together on the day of the exam when their car got a flat tire. On their makeup exam, the instructor asks them to indicate which tire went flat - front left, front right, rear left, or rear right. Supposing that the four students lied and are each choosing a tire at random, what is the probability they each indicate the same tire (and fool their instructor)?

Answer 1

Answer:

The probability they each indicate the same tire is $\frac{1}{64}=0.02$

Step-by-step explanation:

There will be 4 cases involved as there are 4 tires.

Probability of choosing front left tire is $\frac{1}{4}$ as there are a total of 4 tires and we need to choose 1.

Now, if all the four students chose the front left tire then the probability is the product of individual probabilities.

Therefore, probability that all 4 students chooses front left tire is:

$P(fl)=(\frac{1}{4})^4$

Similarly, the probabilities of choosing the remaining 3 tires by all the students would be:

$P(fr)=(\frac{1}{4})^4$, $P(rl)=(\frac{1}{4})^4$, $P(rr)=(\frac{1}{4})^4$

Therefore, the probability that they indicate the same tire is the sum of all these probabilities.

$P(\textrm{same tire by all})=P(fl)+P(fr)+P(rl)+P(rr)\\ P(\textrm{same tire by all})=(\frac{1}{4})^4\times 4\\ P(\textrm{same tire by all})=\frac{4}{4^4}=\frac{1}{4^3}=\frac{1}{64}=0.02$