Where do I find P AUB independent events?

In California, it "never" rains during the summer (one summer when I was there it rained one day every month, and not very hard). If I am planning a picnic, I do not care that it rains one eighth of the days in California; rather that it rains one quarter of the days in September, or one thirtieth of the days in June, depending on when I want my picnic. This is the essence of conditional probability.

The probability of A conditioned on B, denoted P(A|B), is equal to P(AB)/P(B). The division provides that the probabilities of all outcomes within B will sum to 1. Conditioning restricts the sample space to those outcomes which are in the set being conditioned on (in this case B). Note that P(A|B) is not equal to P(B|A); the set after the vertical bar is the set one is conditioning on.

Example: If P(A)=.5, P(B)=.4, and P(AB)=.2 (hence and ), P(A|B)=.2/.4=.5 and P(B|A)=.2/.5=.4.

Example: If P(A)=.5 andP(B|A)=.4, P(BA)=.4 × .5 =.2. (of course AB=BA).

Examples: If P(A)=.5, P(B)=.4, and P(AB)=.2, then P(A|B)=.2/.4=.5 = P(A) and A and B are independent. If P(A)=.6, P(B)=.4, and P(AB)=.2, then P(A|B)=.2/.4=.5 which is not equal to .6=P(A), and A and B are not independent.

• If A and B are independent, P(AB)=P(A)P(B) (because P(A|B)=P(A) for independent events). (Example: If A and B are independent and P(A)=.3 and P(B)=.6, then P(AB)=.3 × .6 = .18.)
• N.B.: If A and B are (which includes the case where A and B are )

Competencies: If P(A)=.5, P(B)=.4, and P(AB)=.3, what is P(A|B)? Are A and B independent?
If P(A)=.6, P(B)=.4, and P(A|B)=.5, what is P(AB)?
If A and B are independent and P(A)=.3, P(B)=.6; P(AB)=?

Reflection: What are the relationships among independence, complementary, and mutually exclusive (disjoint)?

If the probability of one event doesn’t affect the other, you have an independent event. All you do is multiply the probability of one by the probability of another.

Examples

Example 1: The odds of you getting promoted this year are 1/4. The odds of you being audited by the IRS are about 1 in 118. What are the odds that you get promoted and you get audited by the IRS?

Solution:
Step 1: Multiply the two probabilities together:
p(A and B) = p(A) * p(B) = 1/4 * 1/118 = 0.002.
That’s it!

Example 2: The odds of it raining today is 40%; the odds of you getting a hole in one in golf are 0.08%. What are your odds of it raining and you getting a hole in one?

Solution:
Step 1: Multiply the probability of A by the probability of B.
p(A and B) = p(A) * p(B) = 0.4 * 0.0008 = 0.00032.
That’s it!

Formula for the probability of A and B (dependent events): p(A and B) = p(A) * p(B|A)

The formula is a little more complicated if your events are dependent, that is if the probability of one event effects another. In order to figure these probabilities out, you must find p(B|A), which is the conditional probability for the event.

Example question: You have 52 candidates for a committee. Four are persons aged 18 to 21. If you randomly select one person, and then (without replacing the first person’s name), randomly select a second person, what is the probability both people will be between 18 and 21 years old?

Solution:
Step 1: Figure out the probability of choosing an 18 to 21 year old on the first draw. As there are 52 possibilities, and 4 are aged 18 to 21, you have a 4/52 = 1/13 chance.

Step 2: Figure out p(B|A), which is the probability of the next event (choosing a second person aged 18 to 21) given that the first event in Step 1 has already happened.
There are 51 people left, and only 3 are aged 18 to 21 now, so the probability of choosing a young adult again is 3/51 = 1 / 17.

Step 3: Multiply your probabilities from Step 1(p(A)) and Step 2(p(B|A)) together:
p(A) * p(B|A) = 1/13 * 1/17 = 1/221.

Your odds of choosing two people aged 18 to 21 are 1 out of 221.

2. What is the Probability of A or B?

The probability of A or B depends on if you have mutually exclusive events (ones that cannot happen at the same time) or not.

If two events A and B are mutually exclusive, the events are called disjoint events. The probability of two disjoint events A or B happening is:

p(A or B) = p(A) + p(B).

Example question: What is the probability of choosing one card from a standard deck and getting either a Queen of Hearts or Ace of Hearts? Since you can’t get both cards with one draw, add the probabilities:
P(Queen of Hearts or Ace of Hearts) = p(Queen of Hearts) + p(Ace of Hearts) = 1/52 + 1/52 = 2/52.

If the events A and B are not mutually exclusive, the probability is:

(A or B) = p(A) + p(B) – p(A and B).

Example question: What is the probability that a card chosen from a standard deck will be a Jack or a heart?
Solution:

• p(Jack) = 4/52
• p(Heart) = 13/52
• p(Jack of Hearts) = 1/52

So:
p(Jack or Heart) = p(Jack) + p(Heart) – p(Jack of Hearts) = 4/52 + 13/52 – 1/52 = 16/52.

References

Salkind, N. (2019). Statistics for People Who (Think They) Hate Statistics 7th Edition. SAGE.

CITE THIS AS:
Stephanie Glen. "Probability of A and B / A or B" From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/probability-and-statistics/probability-main-index/probability-of-a-and-b/

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