In a school, 3 out of 8 students wear wrist watches. If 5 students are selected at random from the school, find, correct to 2 decimal places, the probability that I) exactly two of them wear wrist watches. II) less than half of them wear wrist watches III) at least three of them wear wrist watches.

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