Math / Data & Statistics

Question5. A modified deck of cards is created by removing the 9 s and 10 s , and the red aces, from a regular deck. a. How many cards are in this deck? (1 mark) b. Given that an 8-card hand has been dealt from this deck, what is the probability of having exactly 3 hearts, 2 diamonds, 2 spades, and 1 club? (3 marks)| c. When drawing two cards from this deck without replacement, what ar odds against getting a pair (two cards of the same rank)? (4 marks)

Studdy Solution

STEP 1

What is this asking? We've got a weird deck of cards, and we want to figure out how many cards are in it, the chances of getting a specific hand, and the odds of not getting a pair when drawing two cards. Watch out! Don't forget that we've removed some cards, so it's not a standard 52-card deck!
Also, probability and odds are different things, so make sure to calculate the right one for part (c).

STEP 2

1. Count the cards
2. Calculate hand probability
3. Calculate odds against a pair

STEP 3

Alright, let's start by figuring out how many cards are removed.
We take out all the 9s and 10s, which means four 9s and four 10s, totaling $4 + 4 = 8$ cards.
We also remove the red aces, which are the heart and diamond aces.
That's two more cards.
So, we've removed a total of $8 + 2 = \mathbf{10}$ cards.

STEP 4

A regular deck has 52 cards.
Since we removed $\mathbf{10}$ cards, our modified deck has $52 - 10 = \mathbf{42}$ cards.

STEP 5

We want to find the probability of getting a very specific 8-card hand from our 42-card deck.
The total number of possible 8-card hands is given by the combination formula: $\binom{42}{8} = \frac{42!}{8!(42-8)!} = \frac{42!}{8!34!} = \frac{42 \cdot 41 \cdot 40 \cdot 39 \cdot 38 \cdot 37 \cdot 36 \cdot 35}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 118,030,143$

STEP 6

Now, let's find the number of ways to get exactly 3 hearts, 2 diamonds, 2 spades, and 1 club.
There are 13 cards in each suit originally, but remember we removed a red ace from hearts and diamonds, and we removed two ranks (9 and 10) from each suit.
This leaves $13 - 1 - 2 = 10$ hearts and diamonds, and $13 - 2 = 11$ spades and clubs.

STEP 7

So, we have $\binom{10}{3}$ ways to choose 3 hearts, $\binom{10}{2}$ ways to choose 2 diamonds, $\binom{11}{2}$ ways to choose 2 spades, and $\binom{11}{1}$ ways to choose 1 club.
Multiplying these together gives us the number of successful outcomes: $\binom{10}{3} \cdot \binom{10}{2} \cdot \binom{11}{2} \cdot \binom{11}{1} = 120 \cdot 45 \cdot 55 \cdot 11 = 3,267,000$

STEP 8

The probability is the number of successful outcomes divided by the total number of possible outcomes: $\frac{3,267,000}{118,030,143} \approx 0.0277$

STEP 9

The odds against an event are the ratio of the number of unfavorable outcomes to the number of favorable outcomes.
First, let's find the number of ways to get a pair.
There are 13 ranks, but two ranks were removed (9 and 10), leaving 11 possible ranks for a pair.
Each rank has four suits, but we removed two aces, so there are only two red aces left.
So, for ranks other than ace, we have $\binom{4}{2}$ ways to get a pair.
For aces, we have $\binom{2}{2}$ ways.
Since there are 10 non-ace ranks and 1 ace rank, the total number of pairs is $10 \cdot \binom{4}{2} + 1 \cdot \binom{2}{2} = 10 \cdot 6 + 1 \cdot 1 = 61$ .

STEP 10

The total number of ways to choose two cards is $\binom{42}{2} = \frac{42 \cdot 41}{2} = 861$ .
The **number of ways to not get a pair** is the total number of ways to choose two cards minus the number of ways to get a pair: $861 - 61 = 800$ .

STEP 11

The odds against getting a pair are the number of ways to not get a pair to the number of ways to get a pair: $800:61$ .

STEP 12

a. There are $\mathbf{42}$ cards in the modified deck. b. The probability of the specified hand is approximately $\mathbf{0.0277}$ . c. The odds against getting a pair are $\mathbf{800:61}$ .

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Studdy Solution

STEP 1

STEP 2

1. Count the cards
2. Calculate hand probability
3. Calculate odds against a pair

STEP 3

STEP 4

A regular deck has 52 cards.
Since we removed $\mathbf{10}$ cards, our modified deck has $52 - 10 = \mathbf{42}$ cards.

STEP 5

STEP 6

STEP 7

STEP 8

The probability is the number of successful outcomes divided by the total number of possible outcomes: $\frac{3,267,000}{118,030,143} \approx 0.0277$

STEP 9

STEP 10

The total number of ways to choose two cards is $\binom{42}{2} = \frac{42 \cdot 41}{2} = 861$ .
The **number of ways to not get a pair** is the total number of ways to choose two cards minus the number of ways to get a pair: $861 - 61 = 800$ .

STEP 11

The odds against getting a pair are the number of ways to not get a pair to the number of ways to get a pair: $800:61$ .

STEP 12

a. There are $\mathbf{42}$ cards in the modified deck. b. The probability of the specified hand is approximately $\mathbf{0.0277}$ . c. The odds against getting a pair are $\mathbf{800:61}$ .

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Studdy Solution

STEP 1

STEP 2

1. Count the cards2. Calculate hand probability3. Calculate odds against a pair

STEP 3

STEP 4

A regular deck has 52 cards.Since we removed 10\mathbf{10}10 cards, our modified deck has 52−10=4252 - 10 = \mathbf{42}52−10=42 cards.

STEP 5

STEP 6

STEP 7

STEP 8

The **probability** is the number of successful outcomes divided by the total number of possible outcomes: 3,267,000118,030,143≈0.0277 \frac{3,267,000}{118,030,143} \approx 0.0277 118,030,1433,267,000​≈0.0277

STEP 9

STEP 10

STEP 11

The **odds against getting a pair** are the number of ways to not get a pair to the number of ways to get a pair: 800:61800:61800:61.

STEP 12

a. There are 42\mathbf{42}42 cards in the modified deck. b. The probability of the specified hand is approximately 0.0277\mathbf{0.0277}0.0277. c. The odds against getting a pair are 800:61\mathbf{800:61}800:61.

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1. Count the cards
2. Calculate hand probability
3. Calculate odds against a pair

A regular deck has 52 cards.
Since we removed $\mathbf{10}$ cards, our modified deck has $52 - 10 = \mathbf{42}$ cards.

The probability is the number of successful outcomes divided by the total number of possible outcomes: $\frac{3,267,000}{118,030,143} \approx 0.0277$

The odds against getting a pair are the number of ways to not get a pair to the number of ways to get a pair: $800:61$ .

a. There are $\mathbf{42}$ cards in the modified deck. b. The probability of the specified hand is approximately $\mathbf{0.0277}$ . c. The odds against getting a pair are $\mathbf{800:61}$ .