Whitney was on the internet researching prices of new and used cars. One website had 17
cars for sale, 13 of which were vans.
If Whitney randomly chose to look at 13 of the cars on the website, what is the probability
that exactly 10 of the chosen cars are vans?
Write your answer as a decimal rounded to four decimal places.



Answer :

To determine the probability that Whitney randomly chose exactly 10 vans out of 13 cars selected from a total of 17 cars (13 of which are vans), we can use the hypergeometric distribution.

The hypergeometric distribution calculates the probability of [tex]\( k \)[/tex] successes (vans, in this context) in [tex]\( n \)[/tex] draws (cars selected), from a finite population of size [tex]\( N \)[/tex] that contains exactly [tex]\( K \)[/tex] successful outcomes (vans).

Here, the parameters are:
- [tex]\( N = 17 \)[/tex] (total cars)
- [tex]\( K = 13 \)[/tex] (total vans)
- [tex]\( n = 13 \)[/tex] (cars selected)
- [tex]\( k = 10 \)[/tex] (vans chosen)

The hypergeometric probability [tex]\( P(X = k) \)[/tex] is given by:
[tex]\[ P(X = k) = \frac{\binom{K}{k} \binom{N - K}{n - k}}{\binom{N}{n}} \][/tex]
Where:
- [tex]\(\binom{a}{b}\)[/tex] is the binomial coefficient "a choose b".

Let's break down the calculations step by step.

1. Calculate the binomial coefficients:
- [tex]\(\binom{K}{k} = \binom{13}{10}\)[/tex]:
[tex]\[ \binom{13}{10} = \frac{13!}{10!(13-10)!} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286. \][/tex]
- [tex]\(\binom{N - K}{n - k} = \binom{4}{3}\)[/tex]:
[tex]\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4. \][/tex]
- [tex]\(\binom{N}{n} = \binom{17}{13}\)[/tex]:
[tex]\[ \binom{17}{13} = \binom{17}{4} \text{(using the property } \binom{n}{k} = \binom{n}{n-k} \text{)} \][/tex]
[tex]\[ \binom{17}{4} = \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1} = 2380. \][/tex]

2. Substitute these values into the hypergeometric probability formula:
[tex]\[ P(X = 10) = \frac{\binom{13}{10} \binom{4}{3}}{\binom{17}{13}} = \frac{286 \times 4}{2380} = \frac{1144}{2380}. \][/tex]

3. Simplify the fraction and convert it to a decimal:
[tex]\[ \frac{1144}{2380} \approx 0.4807. \][/tex]

Therefore, the probability that exactly 10 of the chosen cars are vans is approximately 0.4807 when rounded to four decimal places.