Katyuska selects a playing card at random from the following 666 cards: \{{left brace999 of hearts, 555 of spades, 666 of hearts, 222 of spades, 444 of hearts, 777 of hearts\}}right brace Let AAA be the event that Katyuska selects a spade and BBB be the event that she chooses an odd-numbered card. Which of the following statements are true? Choose all answers that apply: Choose all answers that apply: (Choice A) A P(A\text{ | }B)=P(A)P(A | B)=P(A)P, left parenthesis, A, start text, space, vertical bar, space, end text, B, right parenthesis, equals, P, left parenthesis, A, right parenthesis, the conditional probability that Katyuska selects a spade given that she has chosen an odd-numbered card is equal to the probability that Katyuska selects a spade. (Choice B) B P(B\text{ | }A)=P(B)P(B | A)=P(B)P, left parenthesis, B, start text, space, vertical bar, space, end text, A, right parenthesis, equals, P, left parenthesis, B, right parenthesis, the conditional probability that Katyuska selects an odd-numbered card given that she has chosen a spade is equal to the probability that Katyuska selects an odd-numbered card. (Choice C) C Events AAA and BBB are independent events. (Choice D) D The outcome of events AAA and BBB are dependent on each other. (Choice E) E P(A\text{ and }B)=P(A) \cdot P(B)P(A and B)=P(A)⋅P(B)P, left parenthesis, A, start text, space, a, n, d, space, end text, B, right parenthesis, equals, P, left parenthesis, A, right parenthesis, dot, P, left parenthesis, B, right parenthesis, the probability that Katyuska selects a spade and an odd-numbered card is equal to the probability that Katyuska selects a spade multiplied by the probability that she selects an odd-numbered card.

- Statements A, B, C and E are all true and correct for this question.

- Only Statement D is not true nor correct.

Explanation:

Katyuska wants to select a playing card at random from 6 cards.

- 9 of hearts

- 5 of spades

- 6 of hearts

- 2 of spades

- 4 of hearts

- 7 of hearts

A is the event that Katyuska selects a spade and B is the event that she chooses an odd-numbered card.

Note that the probability of an event is given as the number of elements in that event divided by the Total number of elements in the sample space.

P(A) = n(A) ÷ n(S)

So, of the six cards, there are 2 spades (5 of spades, 2 of spades) and 3 odd numbered cards (9 of hearts, 5 of spades, 7 of hearts)

Probability of event A = Probability of selecting a spade = P(A) = n(A) ÷ n(S)

n(A) = number of spades = 2

n(S) = total number of cards = 6

P(A) = (2/6) = (1/3)

Probability of event B = Probability of selecting an odd numbered card = P(B) = n(B) ÷ n(S)

n(B) = number of odd numbered cards = 3

n(S) = total number of cards = 6

P(B) = (3/6) = (1/2)

P(A n B) = Probability of selecting a spade vard that is also an odd numbered card = n(A n B)

(A n B) = 5 of spades

n(A n B) = 1

n(S) = 6

P(A n B) = (1/6)

So, checking the statements one at a time.

A) Investigate P(A|B) = P(A)

The conditional probability, P(A|B), is given mathematically as

P(A|B) = P(A n B) ÷ P(B)

P(A n B) = (1/6)

P(B) = (1/2)

P(A|B) = (1/6) ÷ (1/2) = (2/6) = (1/3) = P(A)

Hence, this statement is true!

B) Investigate P(B|A) = P(B)

P(B|A) = P(B n A) ÷ P(A)

P(B n A) = P(A n B) = (1/6)

P(A) = (1/3)

P(B|A) = (1/6) ÷ (1/3) = (3/6) = (1/2) = P(B)

This statement is also true!

C) Events A and B are independent events.

Two events are said to be independent when the probability/chances of one of them occurring does not in any way affect the probability/chances of the other one occurring.

Mathematically, the conditions for two events to be independent are

P(A|B) = P(A)

P(B|A) = P(B)

Since these two conditions have already been established, we can conclude that the two events are indeed independent of each other.

D) The outcome of events A and B are dependent on each other.

Since we just proved in (C) that the two events are independent of each other, it necessarily means that this statement is false.

The outcome of events A and B are not dependent on each other.

E) Investigate whether P(A and B) = P(A) × P(B)

P(A) = (1/3)

P(B) = (1/2)

P(A) × P(B) = (1/3) × (1/2) = (1/6) = P(A and B)

This statement is true too.

Hope this Helps!!!

uh, mate. what's the question?