Question

a board game uses the deck of 20 cards shown to the right. two cards are selected at random from this deck. calculate the probability of selecting a bird on the first draw and a lion on the second draw, both with and without replacement.

two cards are to be selected with replacement. determine the probability the first card selected shows a bird and the second shows a lion.
(type an integer or a simplified fraction.)

two cards are to be selected without replacement. determine the probability the first card selected shows a bird and the second shows a lion.
(type an integer or a simplified fraction.)

Explanation:

Analyze the deck composition

We first determine the total number of cards and the count of each category from the given image.

• Total cards: \(20\)
• Row 1 (Birds): \(5\) cards (numbered 1 to 5)
• Row 2 (Squirrels/Acorns): \(5\) cards (numbered 1 to 5)
• Row 3 (Frogs): \(5\) cards (numbered 1 to 5)
• Row 4 (Lions): \(5\) cards (numbered 1 to 5)

Using the Theoretical Probability knowledge point:

• Number of bird cards, \(n(\text{Bird}) = 5\)
• Number of lion cards, \(n(\text{Lion}) = 5\)
• Total number of cards, \(N = 20\)

Calculate probability with replacement

Using the Multiplication Rule for Independent Events knowledge point, the two draws are independent because the first card is replaced before the second draw.

• Probability of drawing a bird first:

$$P(\text{Bird}_1) = \frac{5}{20} = \frac{1}{4}$$

• Probability of drawing a lion second:

$$P(\text{Lion}_2) = \frac{5}{20} = \frac{1}{4}$$

• Combined probability:

$$P(\text{Bird}_1 \text{ and } \text{Lion}_2) = P(\text{Bird}_1) \times P(\text{Lion}_2) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$$

Calculate probability without replacement

Using the Probability without Replacement knowledge point, the outcome of the first draw affects the composition of the deck for the second draw.

• Probability of drawing a bird first:

$$P(\text{Bird}_1) = \frac{5}{20} = \frac{1}{4}$$

• After drawing a bird, \(19\) cards remain in the deck, of which \(5\) are still lions.
• Conditional probability of drawing a lion second:

$$P(\text{Lion}_2 \mid \text{Bird}_1) = \frac{5}{19}$$

• Combined probability:

$$P(\text{Bird}_1 \text{ and } \text{Lion}_2) = P(\text{Bird}_1) \times P(\text{Lion}_2 \mid \text{Bird}_1) = \frac{5}{20} \times \frac{5}{19} = \frac{1}{4} \times \frac{5}{19} = \frac{5}{76}$$

Answer:

Question 1

Two cards are to be selected with replacement. Determine the probability the first card selected shows a bird and the second shows a lion.
<blank>\(\frac{1}{16}\)</blank>

Question 2

Two cards are to be selected without replacement. Determine the probability the first card selected shows a bird and the second shows a lion.
<blank>\(\frac{5}{76}\)</blank>