Answer:
Exactly two of them wear wrist watches: 0.54
Less than half of them wear wrist watches: 0.83
At least three of them wear wrist watches: 0.18
Step-by-step explanation:
To solve this problem, we need to use the concept of the hypergeometric distribution. The hypergeometric distribution describes the probability of [tex] \sf k [/tex] successes in [tex] \sf n [/tex] draws from a finite population of size [tex] \sf N [/tex] that contains exactly [tex] \sf K [/tex] successes, without replacement.
Given:
- Total number of students ([tex] \sf N [/tex]) = 8
- Number of students who wear wrist watches ([tex] \sf K [/tex]) = 3
- Number of students selected ([tex] \sf n [/tex]) = 5
I) Probability that exactly two of them wear wrist watches
We want to find [tex] \sf P(X = 2) [/tex].
The hypergeometric probability formula is:
[tex] \sf P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} [/tex]
Where:
- [tex] \sf \binom{K}{k} [/tex] is the number of ways to choose [tex] \sf k [/tex] successes from [tex] \sf K [/tex] successes.
- [tex] \sf \binom{N-K}{n-k} [/tex] is the number of ways to choose [tex] \sf n-k [/tex] failures from [tex] \sf N-K [/tex] failures.
- [tex] \sf \binom{N}{n} [/tex] is the number of ways to choose [tex] \sf n [/tex] items from [tex] \sf N [/tex] items.
For [tex] \sf k = 2 [/tex]:
[tex] \sf \binom{3}{2} = \frac{3!}{2!(3-2)!} = 3 [/tex]
[tex] \sf \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 [/tex]
[tex] \sf \binom{8}{5} = \frac{8!}{5!(8-5)!} = 56 [/tex]
Thus,
[tex] \sf P(X = 2) = \frac{\binom{3}{2} \binom{5}{3}}{\binom{8}{5}} = \frac{3 \times 10}{56} = \frac{30}{56} = \frac{15}{28} \approx 0.54 [/tex]
II) Probability that less than half of them wear wrist watches
Less than half of 5 is less than 2.5, so we need the probabilities for 0, 1, and 2 wrist watches.
[tex] \sf P(X < 2.5) = P(X = 0) + P(X = 1) + P(X = 2) [/tex]
Calculating for [tex] \sf k = 0 [/tex]:
[tex] \sf \binom{3}{0} = 1 [/tex]
[tex] \sf \binom{5}{5} = 1 [/tex]
Thus,
[tex] \sf P(X = 0) = \frac{\binom{3}{0} \binom{5}{5}}{\binom{8}{5}} = \frac{1 \times 1}{56} = \frac{1}{56} \approx 0.02 [/tex]
Calculating for [tex] \sf k = 1 [/tex]:
[tex] \sf \binom{3}{1} = 3 [/tex]
[tex] \sf \binom{5}{4} = 5 [/tex]
Thus,
[tex] \sf P(X = 1) = \frac{\binom{3}{1} \binom{5}{4}}{\binom{8}{5}} = \frac{3 \times 5}{56} = \frac{15}{56} \approx 0.27 [/tex]
Adding these up:
[tex] \sf P(X < 2.5) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.02 + 0.27 + 0.54 = 0.83 [/tex]
III) Probability that at least three of them wear wrist watches
At least three means three, four, or five.
[tex] \sf P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) [/tex]
Calculating for [tex] \sf k = 3 [/tex]:
[tex] \sf \binom{3}{3} = 1 [/tex]
[tex] \sf \binom{5}{2} = 10 [/tex]
Thus,
[tex] \sf P(X = 3) = \frac{\binom{3}{3} \binom{5}{2}}{\binom{8}{5}} = \frac{1 \times 10}{56} = \frac{10}{56} \approx 0.18 [/tex]
For [tex] \sf k = 4 [/tex] and [tex] \sf k = 5 [/tex], we get 0 because there aren't enough students who wear wrist watches to have 4 or 5 in a selection of 5.
Thus,
[tex] \sf P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \approx 0.18 + 0 + 0 = 0.18 [/tex]
So, the probabilities correct to two decimal places are:
1. Exactly two of them wear wrist watches: 0.54
2. Less than half of them wear wrist watches: 0.83
3. At least three of them wear wrist watches: 0.18
