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So I'm taking a probability class at UNI and we had this question today. it really is not hard but for some reason I can't even figure out how to approach this problem:

There are 50 parking spaces for an event. On average one out of five people going to this event are coming with their own car. The event sold 235 tickets.

What is the probability that all the people who are coming with a car will have space to park their car?

(I'm not actually sure if bayes theorem applies to this problem but I have a feeling it might)

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  • $\begingroup$ have you ever heart about Bernoulli distribution? $\endgroup$ – Seyhmus Güngören Mar 15 '16 at 21:08
  • $\begingroup$ Or of a Binomial distribution. $\endgroup$ – Graham Kemp Mar 15 '16 at 23:54
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Let's rephrase the question:

What is the probability that at most $50$ out of $235$ people will arrive with their cars?

The probability of any single person arriving with his/her car is $\frac15$.

So simply split it into disjoint events, and then add up their probabilities:

  • The probability of exactly $\color\red{0}$ people arriving with their car is $\binom{235}{\color\red{0}}\cdot\left(\frac15\right)^{\color\red{0}}\cdot\left(1-\frac15\right)^{235-\color\red{0}}$
  • The probability of exactly $\color\red{1}$ people arriving with their car is $\binom{235}{\color\red{1}}\cdot\left(\frac15\right)^{\color\red{1}}\cdot\left(1-\frac15\right)^{235-\color\red{1}}$
  • ...
  • The probability of exactly $\color\red{50}$ people arriving with their car is $\binom{235}{\color\red{50}}\cdot\left(\frac15\right)^{\color\red{50}}\cdot\left(1-\frac15\right)^{235-\color\red{50}}$

Hence the answer is:

$$\sum\limits_{n=0}^{50}\binom{235}{n}\cdot\left(\frac15\right)^{n}\cdot\left(1-\frac15\right)^{235-n}\approx71.96\%$$

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HINT 1

Let $X$ be the number of cars arriving. You have to make sure that $X \le 50$. What is the distribution of $X$?

HINT 2

Notice that $X$ is formed by using exactly the same question (are you driving?) with 2 outcomes (yes and no) for each of the 235 guests. What is the probability an average guest will answer "yes"?

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