Chapter 3: Q. 3.25 (page 203)

In a standard deck, there are $52$ cards. Twelve cards are face cards (F) and $40$ cards are not face cards (N). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.
a. Find P(FN OR NF).
b. Find P(NF).
c. Find P (at most one face card).
Hint: "At most one face card" means zero or one face card.
d. Find P (at least on face card).
Hint: "At least one face card" means one or two face cards.

Short Answer

Expert verified

a. The value of P(F N O R N F) is $0.362$ .

b. The value of P(N F) is $0.784$ .

c. The value of P (at most one face card) is $0.950$ .

d. The value of P (at least on face card) is $0.412$ .

Step by step solution

Introduction

Number of cards in a deck = $52$

face cards = $12$ and non face cards = $40$

Explanation (Part a)

We have a tree diagram.

As a result, the needed probability can be computed as follows:

$P (F N O R N F) = P (F N) + P (N F) = \frac{480}{2562} + \frac{480}{2562} = 0.362$

Explanation (Part b)

We have a tree diagram.

As a result, the needed probability can be computed as follows:

$P (N ∣ F) = \frac{40}{51} = 0.784$

Explanation (Part c)

We have a tree diagram.

As a result, the needed probability can be computed as follows:

$P (at most one face card) = P (F N) + P (N F) + P (N N) = \frac{480}{2652} + \frac{480}{2652} + \frac{1560}{2652} = 0.950$

Explanation (Part d)

We have a tree diagram.

As a result, the needed probability can be computed as follows:

$P (at least one face card) = P (F F) + P (F N) + P (N F) = \frac{132}{2652} + \frac{480}{2652} + \frac{480}{2652} = 0.412$

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Short Answer

Step by step solution

Introduction

Explanation (Part a)

Explanation (Part b)

Explanation (Part c)

Explanation (Part d)

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