Probability Combined events

 Probability of combined events refers to the likelihood of two or more events occurring together. It involves calculating the probability of each individual event and then combining them using specific rules.

There are two main rules used to calculate probabilities of combined events: the Addition Rule and the Multiplication Rule.

1. Addition Rule:

The Addition Rule is used to calculate the probability of either of two events occurring. It states that the probability of either Event A or Event B happening is equal to the sum of their individual probabilities minus the probability of both events happening simultaneously.

P(A or B) = P(A) + P(B) - P(A and B)

Example: Let's say you have a bag of marbles containing 5 red marbles and 7 blue marbles. What is the probability of drawing either a red or a blue marble?

P(Red) = 5/12

P(Blue) = 7/12

P(Red or Blue) = P(Red) + P(Blue) = 5/12 + 7/12 = 12/12 = 1

So, the probability of drawing either a red or a blue marble is 1 or 100%.

2. Multiplication Rule:

The Multiplication Rule is used to calculate the probability of two or more independent events occurring simultaneously. It states that the probability of both Event A and Event B happening is equal to the product of their individual probabilities.

P(A and B) = P(A) * P(B)

Example: Let's say you flip a fair coin twice. What is the probability of getting two heads?

P(Head) = 1/2

P(Head and Head) = P(Head) * P(Head) = 1/2 * 1/2 = 1/4

So, the probability of getting two heads is 1/4 or 25%.

These rules can be extended to more than two events by continuously applying the Addition Rule or the Multiplication Rule, depending on the situation.

3. Conditional Probability:

Conditional probability refers to the probability of an event A occurring given that another event B has already occurred. It is denoted as P(A|B) and is calculated by dividing the probability of both events A and B occurring by the probability of event B.

P(A|B) = P(A and B) / P(B)

Example: Consider a deck of 52 playing cards. What is the probability of drawing a heart card given that a red card has already been drawn?

P(Heart | Red) = P(Heart and Red) / P(Red)

               = (13/52) / (26/52)

               = 13/26

               = 1/2

So, the probability of drawing a heart card given that a red card has been drawn is 1/2 or 50%.

4. Independent Events:

Events are considered independent if the occurrence or non-occurrence of one event does not affect the probability of the other event. If two events, A and B, are independent, then the probability of both events occurring is equal to the product of their individual probabilities.

P(A and B) = P(A) * P(B)

Example: You roll a fair six-sided die and flip a fair coin. What is the probability of rolling a 4 and getting heads on the coin flip?

P(Rolling a 4) = 1/6

P(Heads) = 1/2

P(Roll a 4 and Heads) = P(Rolling a 4) * P(Heads)

                      = 1/6 * 1/2

                      = 1/12

So, the probability of rolling a 4 and getting heads is 1/12.

These are some key concepts related to probability and combined events. By understanding these concepts and applying the appropriate rules, you can calculate probabilities for complex situations involving multiple events.

Remember, practice and real-life examples are essential for developing a deeper understanding of probability and its application in various fields such as statistics, decision-making, and risk analysis.

If you have any more specific questions or need further explanations, feel free to ask on the comments section!

Here are some questions on probability of combined events along with sample answers:

1. Question: A bag contains 4 red balls and 6 blue balls. If two balls are drawn without replacement, what is the probability of drawing a red ball followed by a blue ball?

   Sample Answer: The probability of drawing a red ball first is 4/10. After one red ball is removed, there are 9 balls left in the bag, including 6 blue balls. Therefore, the probability of drawing a blue ball is 6/9. To find the probability of both events happening, we multiply the individual probabilities: (4/10) * (6/9) = 24/90 = 4/15.

2. Question: A deck of cards contains 52 cards. If two cards are drawn in succession without replacement, what is the probability of drawing two kings?

   Sample Answer: The probability of drawing the first king is 4/52, as there are four kings in the deck. After one king is removed, there are 51 cards left, including three kings. Therefore, the probability of drawing the second king is 3/51. To find the probability of both events happening, we multiply the individual probabilities: (4/52) * (3/51) = 12/2652 = 1/221.

3. Question: A bowl contains 8 red marbles, 5 blue marbles, and 7 green marbles. If three marbles are drawn without replacement, what is the probability of drawing a blue marble, a red marble, and then a green marble?

   Sample Answer: The probability of drawing a blue marble first is 5/20. After one blue marble is removed, there are 19 marbles left, including 8 red marbles. Therefore, the probability of drawing a red marble is 8/19. After one red marble is removed, there are 18 marbles left, including 7 green marbles. Therefore, the probability of drawing a green marble is 7/18. To find the probability of all three events happening, we multiply the individual probabilities: (5/20) * (8/19) * (7/18) = 280/6840 = 5/114.

4. Question: A box contains 10 red socks, 8 blue socks, and 5 green socks. If two socks are drawn at random without replacement, what is the probability of drawing a red sock followed by a blue sock?

   Sample Answer: The probability of drawing a red sock first is 10/23. After one red sock is removed, there are 22 socks left in the box, including 8 blue socks. Therefore, the probability of drawing a blue sock is 8/22. To find the probability of both events happening, we multiply the individual probabilities: (10/23) * (8/22) = 80/506 = 40/253.

5. Question: A bag contains 6 white balls and 4 black balls. If three balls are drawn one at a time without replacement, what is the probability of drawing two white balls and then a black ball?

   Sample Answer: The probability of drawing a white ball first is 6/10. After one white ball is removed, there are 9 balls left in the bag, including 5 white balls. Therefore, the probability of drawing a second white ball is 5/9. After two white balls are removed, there are 8 balls left, including 4 black balls. Therefore, the probability of drawing a black ball is 4/8. To find the probability of all three events happening, we multiply the individual probabilities: (6/10) * (5/9) * (4/8) = 120/720 = 1/6.

6. Question: A box contains 15 identical keys, of which 4 are faulty. If three keys are randomly selected with replacement, what is the probability of selecting two faulty keys and then a working key?

   Sample Answer: The probability of selecting a faulty key is 4/15, and since the selection is made with replacement, this probability remains the same for each selection. Therefore, the probability of selecting two faulty keys in a row is (4/15) * (4/15) = 16/225. The probability of selecting a working key is 1-4/15 = 11/15, since there are 11 non-faulty keys. To find the probability of the given sequence of events, we multiply the individual probabilities: (16/225) * (16/225) * (11/15) = 2816/759375.

These additional examples should give you more practice in calculating probabilities for combined events. Remember to adjust the probabilities based on the conditions of the problem and always consider whether the selections are made with or without replacement.