Question

Suppose a deck of cards contains 13 cards:

Answer 1

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

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

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