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1. Michelle used a standard deck of 52 cards and selected a card at random. Afterrecording the suit of the card she picked, she then replaced the card.SultOutcomeSpades 9Hearts11Clubs7Diamonds3Part A: Determine the empirical probability of selecting a heart.Part B: Determine the theoretical probability of selecting a heart.Part C: Determine the empirical probability of selecting a club or diamond.Part D: Determine the theoretical probability of selecting a club or diamond.

Sagot :

The card deck is a standard deck of 52 cards, with 13 cards in each suit.

For the experiment carried out by Michelle, we have the following information:

n(Spades) = 9

n(Hearts) = 11

n(Clubs) = 7

n(Diamonds) = 3

The total number of times she performed the experiment is

[tex]n(\text{Total) = 9+11+7+3 = 30}[/tex]

The empirical probability will make use of the experimental results, while the theoretical probability will make use of the total possibilities.

PART A: Empirical Probability of selecting a heart.

Probability is calculated by

[tex]P(\text{outcome) = }\frac{n(outcome)}{n(total)}[/tex]

Therefore, the probability is calculated as

[tex]P(\text{heart) = }\frac{11}{30}[/tex]

PART B: Theoretical probability of selecting a heart.

This is calculated by

[tex]\begin{gathered} P(\text{heart) = }\frac{13}{52} \\ P(\text{heart) = }\frac{1}{4} \end{gathered}[/tex]

PART C: Empirical probability of selecting a club or diamond.

To calculate the probability for two outcomes, A or B, the probability can be calculated by

[tex]P(A\text{ or B) = P(A) + P(B)}[/tex]

Therefore, we will find the probability of getting a club and then a diamond.

[tex]P(\text{heart) = }\frac{11}{30}[/tex][tex]P(\text{diamond) = }\frac{3}{30}=\frac{1}{10}[/tex]

Therefore, the probability of selecting a club or a diamond is

[tex]\begin{gathered} P(\text{heart or diamond) = }\frac{11}{30}+\frac{1}{10} \\ P(\text{heart or diamond) = }\frac{7}{15} \end{gathered}[/tex]

PART D: Theoretical probability of selecting a club or a diamond

We will find the probability of getting a club and then a diamond.

[tex]P(\text{club) = }\frac{13}{52}=\frac{1}{4}[/tex][tex]P(\text{diamond) = }\frac{13}{52}=\frac{1}{4}[/tex]

Therefore, the probability of selecting a club or a diamond is

[tex]\begin{gathered} P(\text{heart or diamond) = }\frac{1}{4}+\frac{1}{4} \\ P(\text{heart or diamond) = }\frac{1}{2} \end{gathered}[/tex]