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A box contains 13 chips, of which 5 are blue and the remaining are red. We selected 3 chips at random. Let [tex]X[/tex] denote the number of blue chips in our selection.

(a) [tex]X[/tex] follows a Hyper-geometric distribution.
(b) [tex]P(X = 3) = \qquad 0.035[/tex]
(c) [tex]P(X = 0) = \qquad 0.196[/tex]
(d) [tex]P(X = 1) = \qquad 0.4895[/tex]
(e) [tex]P(X \leq 1) = \qquad[/tex]
(f) [tex]P(X \ \textgreater \ 1) = \qquad[/tex]

Sagot :

GinnyAnsweridn
Let's tackle this question step by step, beginning from the top and explaining the concepts as we go along.

### Given Information:
- Total number of chips = 13
- Number of blue chips = 5
- Number of chips selected = 3

Let [tex]\( X \)[/tex] be the number of blue chips in our selection.

### (a)
Since we are sampling without replacement from a finite population, [tex]\( X \)[/tex] follows a Hypergeometric distribution. We can write this as:

[tex]\[ X \sim \text{Hypergeometric}(M, n, N) \][/tex]

where:
- [tex]\( M \)[/tex] is the population size (13 chips total).
- [tex]\( n \)[/tex] is the number of success states in the population (5 blue chips).
- [tex]\( N \)[/tex] is the number of draws (3 chips drawn).

So, [tex]\( X \)[/tex] follows a Hypergeometric distribution with parameters 13, 5, and 3.

### (b)
We need to calculate the probability of drawing exactly 3 blue chips, denoted as [tex]\( P(X = 3) \)[/tex].

[tex]\[ P(X = 3) \approx 0.035 \][/tex]

### (c)
We need to find the probability of drawing 0 blue chips, denoted as [tex]\( P(X = 0) \)[/tex].

[tex]\[ P(X = 0) \approx 0.196 \][/tex]

### (d)
We need the probability of drawing exactly 1 blue chip, denoted as [tex]\( P(X = 1) \)[/tex].

[tex]\[ P(X = 1) \approx 0.4895 \][/tex]

### (e)
Now, the probability of drawing 1 or fewer blue chips (i.e., [tex]\( X \)[/tex] being less than or equal to 1). This is the sum of the probabilities of [tex]\( X = 0 \)[/tex] and [tex]\( X = 1 \)[/tex]:

[tex]\[ P(X \leq 1) = P(X = 0) + P(X = 1) \approx 0.196 + 0.4895 \approx 0.6853 \][/tex]

### (f)
Finally, the probability of drawing more than 1 blue chip (i.e., [tex]\( X \)[/tex] being greater than 1). This can be found by subtracting the probability of [tex]\( X \leq 1 \)[/tex] from 1.

[tex]\[ P(X > 1) = 1 - P(X \leq 1) \approx 1 - 0.6853 \approx 0.3147 \][/tex]

### Summary:
(a) [tex]\( X \)[/tex] follows a Hypergeometric distribution.

(b) [tex]\( P(X = 3) \approx 0.035 \)[/tex]

(c) [tex]\( P(X = 0) \approx 0.196 \)[/tex]

(d) [tex]\( P(X = 1) \approx 0.4895 \)[/tex]

(e) [tex]\( P(X \leq 1) \approx 0.6853 \)[/tex]

(f) [tex]\( P(X > 1) \approx 0.3147 \)[/tex]
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