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irina1246 [14]
2 years ago
13

Suppose a deck of cards contains 13 cards:

Mathematics
1 answer:
ira [324]2 years ago
7 0

Answer:

P(G_1) = \frac{5}{13}

P(G_2) = \frac{1}{3}

P(X \ge 1) = \frac{25}{39}

Step-by-step explanation:

Given

G = 5

R = 4

B = 4

n = 13

Solving (a): P(G_1)

This is calculated as:

P(G_1) = \frac{G}{n}

P(G_1) = \frac{5}{13}

Solving (b): P(G_2)

This is calculated as:

P(G_2) = \frac{G - 1}{n - 1} -- this is so because the selection is without replacement

P(G_2) = \frac{5 - 1}{13 - 1}

P(G_2) = \frac{4}{12}

P(G_2) = \frac{1}{3}

Solving (c): P(X \ge 1)

Using the complement rule, we have:

P(X \ge 1) = 1 - P(X = 0)

To calculate P(X = 0), we have:

G = 5 --- Green

G' = 8 ---- Not green

The probability that both selections are not green is:

P(X = 0) = P(G'_1) * P(G'_2)

So, we have:

P(X = 0) = \frac{G'}{n} * \frac{G'-1}{n-1}

P(X = 0) = \frac{8}{13} * \frac{8-1}{13-1}

P(X = 0) = \frac{8}{13} * \frac{7}{12}

Simplify

P(X = 0) = \frac{2}{13} * \frac{7}{3}

P(X = 0) = \frac{14}{39}

Recall that:

P(X \ge 1) = 1 - P(X = 0)

P(X \ge 1) = 1 - \frac{14}{39}

Take LCM

P(X \ge 1) = \frac{39 -14}{39}

P(X \ge 1) = \frac{25}{39}

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Given:
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Refer to the diagram shown below.
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Also,
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Answer:
The coordinates of D are (1.11, 1)

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