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.
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.
