9-6: Properties of Probability

This section is basically taking the formulas we already learned (for "and" and "or" situations) and applying them to probabilities of certain outcomes, rather than just numbers of ways those outcomes can occur.

Remember that P(E) = n(E)/n(S)

The most important thing to remember in this section is to carefully assess whether you are dealing with an "and" or "or" situation.

AND (multiply *) a.k.a. Intersection of Events A∩B

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

These are independent events, which means the way one occurs does not affect the way the other could occur.

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

These are non-independent events, so the formula is the probability of A happening times the probability of B happening given that A has already occured. You must take into account the limited choices you have for event B after event A has happened.

OR (add +) a.k.a. Union of Events AUB
P(A or B) = P(A) + P(B)

These are mutually exclusive events, which means they do not overlap.

P(A or B) = P(A) + P(B) - P(A∩B)

These are non-mutually exclusive events. Somewhere, there is an overlap in possible outcomes. You must subtract this overlap.

Complementary Events
This is pretty much common sense.

P(not A) = 1 - P(A)

The probability that event A will NOT occur is 1 minus the probability that event A WILL occur. This makes sense because P(A) and P(not A) cover all possibilities, which is why their sum is 1, or 100%.

This also relates to...
P(at least 1) = 1 - P(none)
P(at least 2) = 1 - P(at most 1)
P(at least 3) = 1 - P(at most 2)
etc...

Example 1
The Mad Hatter and March Hare are having tea again. Every time they get caught up in song, they break their dishes because they're not paying attention. The probability of breaking a teacup is 70%. The probability of breaking a tea saucer is 40%. The probability of breaking a tea kettle is 20%.

What is the probability that...
a. They break one of each?
b. They break none of them?
c. They break only a teacup (not a saucer or kettle)?
d. They break a saucer or a teacup?
e. They break a kettle and teacup but not a saucer?

Solution
a. P(teacup) * P(saucer) * P(kettle)
You multiply, since they are breaking a teacup AND a saucer AND a kettle.

.7 * .4 * .2 = .056 = 5.6%

b. The probability of NOT breaking something is equal to 1 minus the probability of breaking something.

.3 * .6 * .8 = .144 = 14.4%

c. P(only a teacup) = P (teacup) * P(not saucer) * P(not kettle)

.7 * .6 * .8 = .336 = 33.6%

d. P(saucer or teacup) = P(saucer) + P(teacup) - P(saucer and teacup)
Be careful! These are NOT mutually exclusive events, since they could break both a saucer AND a teacup.

(.4 + .7) - (.4*.7) = .82 = 82%

e. P(kettle) * P(teacup) * P(not saucer)

.2 * .7 * .6 = .084 = 8.4%
Example 2
Alice has a 77% chance of winning a game of crochet. The Queen of Hearts has a 98% chance (since she cheats). Tweedledee and Tweedledum have a 4% chance of winning.

If they all play in separate games, what is the probability that...
a. Alice, the Queen, and the twins win?
b. At least one of them wins?

Solution
a. P(Alice) * P(Queen) * P(Twins) = .77 * .98 * .04 = .030184 = 3.0184%

b. P(at least 1) = 1 - P(none)
= 1 - (.23 * .02 * .96)
= 1 - .004416
= .995584 = 99.5584%

Extra Help
Probability - Summary, Formulas, Examples
If you're confused, this is a great site to look at!

And...

high-five anyone? =)

Christina, you're up next!