Analyzing Card Game Outcomes
Analyzing Card Game Outcomes
(Analyzing Card Game Outcomes)
it’s about big data assingmnt please see the question below
Consider the experiment of drawing two cards without replacement from a deck consisting of only the Ace through 10 of a single suit (e.g. only hearts).
- Describe the outcomes of this experiment. List the elements of the sample space.
- Define the event Ai to be the set of outcomes for which the sum of values of the cards is I (with an Ace = 1). List the outcomes associated with Ai for i=3 to 19.
- What is the probability of obtaining a sum of the two cards equaling from 3 to 19?
- Now let event B the event “total card value is odd).†Find P(B) and P(Bc)
- What is the probability that the sum of two cards will be more than 14)
(You can use Excel If you want to use it as a tool to organize your answer sheet.)
(Analyzing Card Game Outcomes)
1. Outcomes of the Experiment and Sample SpaceThe sample space consists of all possible pairs of cards drawn from the deck. Since there are 10 cards (Ace to 10), the number of ways to choose 2 cards from 10 is calculated as follows:
Total outcomes=(102)=10×92=45\text{Total outcomes} = \binom{10}{2} = \frac{10 \times 9}{2} = 45The sample space SS can be represented as all combinations of the cards:
S={(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(1,9),(1,10),(2,3),(2,4),(2,5),(2,6),(2,7),(2,8),(2,9),(2,10),(3,4),(3,5),(3,6),(3,7),(3,8),(3,9),(3,10),(4,5),(4,6),(4,7),(4,8),(4,9),(4,10),(5,6),(5,7),(5,8),(5,9),(5,10),(6,7),(6,8),(6,9),(6,10),(7,8),(7,9),(7,10),(8,9),(8,10),(9,10)}S = \{(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (3, 10), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (4, 10), (5, 6), (5, 7), (5, 8), (5, 9), (5, 10), (6, 7), (6, 8), (6, 9), (6, 10), (7, 8), (7, 9), (7, 10), (8, 9), (8, 10), (9, 10)\}2. Define Event AiA_iEvent AiA_i is defined as the set of outcomes for which the sum of the values of the cards equals ii. The possible sums range from 3 (Ace + Ace) to 19 (10 + 9). Below are the outcomes associated with each ii from 3 to 19.
- A3A_3: {(1, 2)}
- A4A_4: {(1, 3)}
- A5A_5: {(1, 4), (2, 3)}
- A6A_6: {(1, 5), (2, 4)}
- A7A_7: {(1, 6), (2, 5), (3, 4)}
- A8A_8: {(1, 7), (2, 6), (3, 5)}
- A9A_9: {(1, 8), (2, 7), (3, 6), (4, 5)}
- A10A_{10}: {(1, 9), (2, 8), (3, 7), (4, 6)}
- A11A_{11}: {(1, 10), (2, 9), (3, 8), (4, 7), (5, 6)}
- A12A_{12}: {(2, 10), (3, 9), (4, 8), (5, 7)}
- A13A_{13}: {(3, 10), (4, 9), (5, 8), (6, 7)}
- A14A_{14}: {(4, 10), (5, 9), (6, 8), (7, 7)}
- A15A_{15}: {(5, 10), (6, 9), (7, 8)}
- A16A_{16}: {(6, 10), (7, 9)}
- A17A_{17}: {(7, 10)}
- A18A_{18}: {(8, 10)}
- A19A_{19}: {(9, 10)}
3. Probability of Obtaining a Sum from 3 to 19The total number of outcomes is 45. We can calculate the probabilities P(Ai)P(A_i) for each ii from 3 to 19 by dividing the number of outcomes in each AiA_i by the total outcomes (45). Here are the probabilities for sums from 3 to 19:
- P(A3)=145P(A_3) = \frac{1}{45}
- P(A4)=145P(A_4) = \frac{1}{45}
- P(A5)=245P(A_5) = \frac{2}{45}
- P(A6)=245P(A_6) = \frac{2}{45}
- P(A7)=345=115P(A_7) = \frac{3}{45} = \frac{1}{15}
- P(A8)=345=115P(A_8) = \frac{3}{45} = \frac{1}{15}
- P(A9)=445P(A_9) = \frac{4}{45}
- P(A10)=445P(A_{10}) = \frac{4}{45}
- P(A11)=545=19P(A_{11}) = \frac{5}{45} = \frac{1}{9}
- P(A12)=445P(A_{12}) = \frac{4}{45}
- P(A13)=445P(A_{13}) = \frac{4}{45}
- P(A14)=445P(A_{14}) = \frac{4}{45}
- P(A15)=345=115P(A_{15}) = \frac{3}{45} = \frac{1}{15}
- P(A16)=245P(A_{16}) = \frac{2}{45}
- P(A17)=145P(A_{17}) = \frac{1}{45}
- P(A18)=145P(A_{18}) = \frac{1}{45}
- P(A19)=145P(A_{19}) = \frac{1}{45}
4. Event BB: Total Card Value is OddEvent BB occurs when the sum of the two cards is odd. To determine P(B)P(B) and P(Bc)P(B^c):
- Pairs yielding odd sums:
The pairs that lead to an odd sum are:- (1, 2) → 3
- (1, 4) → 5
- (1, 6) → 7
- (1, 8) → 9
- (1, 10) → 11
- (2, 3) → 5
- (2, 5) → 7
- (2, 7) → 9
- (2, 9) → 11
- (3, 4) → 7
- (3, 6) → 9
- (3, 8) → 11
- (4, 5) → 9
- (4, 7) → 11
- (5, 6) → 11
- (5, 8) → 13
- (6, 7) → 13
- (6, 9) → 15
- (7, 8) → 15
- (7, 10) → 17
- (8, 9) → 17
- (9, 10) → 19
Counting these outcomes, there are 23 outcomes resulting in odd sums.
Thus,
P(B)=2345P(B) = \frac{23}{45} P(Bc)=1−P(B)=2245P(B^c) = 1 – P(B) = \frac{22}{45}5. Probability That the Sum of Two Cards is More Than 14To find the probability that the sum is more than 14, we list the pairs that result in sums greater than 14:
- A15A_{15}: {(5, 10), (6, 9), (7, 8)} → 3 outcomes
- A16A_{16}: {(6, 10), (7, 9)} → 2 outcomes
- A17A_{17}: {(7, 10)} → 1 outcome
- A18A_{18}: {(8, 10)} → 1 outcome
- A19A_{19}: {(9, 10)} → 1 outcome
Thus, the total number of outcomes for sums greater than 14 is:
3+2+1+1+1=83 + 2 + 1 + 1 + 1 = 8So, the probability that the sum is greater than 14 is:
P(sum>14)=P(\text{sum} > 14) =
- Total Outcomes: 45
- P(Ai)P(A_i) for i=3i = 3 to 1919: As listed above
- P(B)=2345P(B) = \frac{23}{45}, P(Bc)=2245P(B^c) = \frac{22}{45}
- P(sum>14)=845P(\text{sum} > 14) = \frac{8}{45}
