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*label*Algebra

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1. Give an example of an independent event.

2. Give an example of a dependent event

3. What is the formula used to calculate the probability of an independent event?

4. What is the formula used to calculate the probability of dependent event?

Q.1) AN EXAMPLE OF AN INDEPENDENT EVENT IS TOSS OF A COIN

Q.2)EXAMPLE OF AN DEPENDENT EVENT MAY BE A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen.The probability that the first card is a queen is 4 out of 52. However, if the first card is not replaced, then the second card is chosen from only 51 cards. Accordingly, the probability that the second card is a jack given that the first card is a queen is 4 out of 51.The outcome of choosing the first card has affected the outcome of choosing the second card, making these events dependent.

Q.3) FORMULA USED FOR INDEPENDENT EVENT WITH EXAMPLES : You are playing a game that involves spinning the money wheel shown. During your turn you get to spin the wheel twice. What is the probability that you get more than $500 on your first spin and then go bankrupt on your second spin? SOLUTION Let event A be getting more than $500 on the first spin, and let event B be going bankrupt on the second spin. The two events are independent. So, the probability is: P(A and B) = P(A) • P(B) = 2 8 4 • 2 2 4 = 3 1 6 ≈ 0.028 . .

Q.4) FORMULA USED FOR DEPENDENT EVEN WITH AN EXAMPLE : You randomly select two cards from a standard 52-card deck. What is the probability that the first card is not a face card (a king, queen, or jack) and the second card is a face card if (a) you replace the first card before selecting the second, and (b) you do not replace the first card? SOLUTION a. If you replace the first card before selecting the second card, then A and B are independent events. So, the probability is: P(A and B) = P(A) • P(B) = 4 5 0 2 • 1 5 2 2 = 1 3 6 0 9 ≈ 0.178 b. If you do not replace the first card before selecting the second card, then A and B are dependent events. So, the probability is: P(A and B) = P(A) • P(B|A) = 4 5 0 2 • 1 5 2 1 = 2 4 2 0 1 ≈ 0.181 EXAMPLE 6

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