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    singh2010's Avatar
    singh2010 Posts: 13, Reputation: 1
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    #1

    Apr 28, 2010, 11:25 AM
    probability
    I have four formulas for probability. They are:

    P (A or B) = P(A) + P(B) - Mutually exclusive
    P(A1), P(A2),. P(An) = 1 - Mutually exclusive and exhaustive
    P (A and B) = P(A) x P(B) - Independent
    P (A and B) = P(A) x P(B|A) - Conditional probability

    I have all these formulas but when I look at questions they all seem to be the same for each formula, I am having trouble trying to know what kind of formula I use for any given question.

    I would be very appreciative if someone can give me an example of a question for each probability. Thanks
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Apr 28, 2010, 01:20 PM

    Sure - here you go:

    1. Mutally exclusive: P(A or B) = P(A) + P(B) . Mutually exclusive means either A can happen, or B can happen, or maybe neither happens, but they can't both happen. Example: in rolling a dice let P(A) = probability of rolling a 1 and P(B) = probability of rolling a 2. I trust you can see that P(A) = 1/6, and P(B) = 1/6, because the chances of rolling any particular number is 1 out of 6. Now, the probability of rolling either a 1 or a 2 is:

    P(A or B) = P(A) + P(B) = 1/6 +1/6 = 1/3.

    Note that rolling a 1 or a 2 is mutually exclusive - you can't roll both at the same time with one dice.

    2. Mutually exclusive and exhaustive: means you provide all possible outcomes for mutually exclusive events. Example: in rolling a dice you can get a 1, 2, 3, 4, 5, or 6. That is the complete list of possible outcomes. The probability of any of these outcomes occurring is 1/6. So P(A+B+... +F) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1.

    3. Independent: P(A and B) = P(A) x P(B). Example - the probability of roling a dice and getting a 1 is P(A) = 1/6. The probability of flipping a coin and getting a heads is P(B) = 1/2. Clearly the outcome of the dice has no bearing on the flip of the coin, so these two events are independent of each other. Hence if you roll the dice and flip the coin, the probability of rolling a 1 and flipping for a heads is 1/6 x 1/2 = 1/12.

    4. Conditional probability: P(A and B) = P(A) x P (B|A). This is used for non-independent events, such as: in making two draws from a deck of cards, what is the probability of drawing first an Ace and then a King, in that order. Here P(A) = probability of drawing an ace, which is 4/52. P(B|A) is the probability of drawing a King given that you already drew an Ace. In other words, from the original deck of 52 cards you've already removed one ace, so the probability now of drawing a King is 4/51. Hence the probability of draiwnag an ace followed by a King is P(A and B) = P(A) x P(B|A) = 4/52 * 4/51.

    Hope this helps!
    singh2010's Avatar
    singh2010 Posts: 13, Reputation: 1
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    #3

    Apr 29, 2010, 05:58 AM
    Quote Originally Posted by ebaines View Post
    4. Conditional probability: P(A and B) = P(A) x P (B|A). This is used for non-independent events, such as: in making two draws from a deck of cards, what is the probability of drawing first an Ace and then a King, in that order. Here P(A) = probability of drawing an ace, which is 4/52. P(B|A) is the probability of drawing a King given that you already drew an Ace. In other words, from the original deck of 52 cards you've already removed one ace, so the probability now of drawing a King is 4/51. Hence the probability of draiwnag an ace followed by a King is P(A and B) = P(A) x P(B|A) = 4/52 * 4/51.
    So for this last one the answer is 4/52 * 4/51 or do I have to times them and get the answer?
    The question I have for this probability is two households go to buy coffee, if one neighbour buys a particular brand there is a 60% chance the second housefold will copy them.
    There is a 80% chance the first household will buy that coffee. What is the probability that both households buy that coffee?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Apr 29, 2010, 06:42 AM
    Quote Originally Posted by singh2010 View Post
    So for this last one the answer is 4/52 * 4/51 or do i have to times them and get the answer?
    Sorry - I used an asterisk to mean multiplication when it would have been clearer if I had used the "x" mark instead. So it's 4/52 x 4/51.

    Quote Originally Posted by singh2010 View Post
    The question i have for this probability is two households go to buy coffee, if one neighbour buys a particular brand there is a 60% chance the second housefold will copy them.
    There is a 80% chance the first household will buy that coffee. What is the probability that both households buy that coffee?
    This is conditional probability: P(A and B) = P(A) x P(B|A). Here P(A) = 80%, and P(B|A) is 60%. Can you take it from here?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Apr 29, 2010, 09:10 AM

    Right ebaines, this is conditional probability. Often, singh2010, you'll notice the word "if" or "given". Those two words often indicate a conditional probability.

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