Download Chapter 5.3 Conditional Probability

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Transcript
More of David Palay’s
Slides
Conditional Probability
Given that you had a quiz on Tuesday,
what is the Probability that you have
homework tonight…
Example:
Suppose one employee is selected from a group of
random employees. We’re interested if this
employee has used the proper cover sheet for his
TPS report (since there’s a new cover sheet and
we’re not certain that this employee has read the
memo).
We could ask:
• What is 𝑃(𝑖𝑚𝑝𝑟𝑜𝑝𝑒𝑟 𝑇𝑃𝑆)
We could ALSO ask,
• What is 𝑃(𝑖𝑚𝑝𝑟𝑜𝑝𝑒𝑟 𝑇𝑃𝑆 | 𝑚𝑎𝑙𝑒)
𝑃(𝑖𝑚𝑝𝑟𝑜𝑝𝑒𝑟 𝑇𝑃𝑆 | 𝑀𝑎𝑙𝑒)
• This is the probability of the employee’s TPS
report having the wrong coversheet, GIVEN
that the employee is male.
Conditional Probability
• Written as
𝑃(𝐵|𝐴)
• Read as
Probability of event B, given event A has already
happened.
Our Company
• There are 15 people in the small company.
• 9 are male
• 5 females filed improper TPS reports
• 8 employees filed improper TPS reports.
• From this, we can figure out the probability of
having an improper TPS report given that the
employee is male.
Method 1
• Create a table:
Male
Female
Proper TPS
Improper TPS
Total
• Use table to calculate probability
Total
Example
• When rolling 2d6, what is the probability that the
sum is a 7, GIVEN THAT ONE OF THE DICE IS A 1.
{1,1} {1,2} {1,3} {1,4} {1,5} {1,6}
{2,1} {2,2} {2,3} {2,4} {2,5} {2,6}
{3,1} {3,2} {3,3} {3,4} {3,5} {3,6}
{4,1} {4,2} {4,3} {4,4} {4,5} {4,6}
{5,1} {5,2} {5,3} {5,4} {5,5} {5,6}
{6,1} {6,2} {6,3} {6,4} {6,5} {6,6}
Method 2: Calculating
𝑃 𝐴∩𝐵
𝑃 𝐵𝐴 =
𝑃 𝐴
TPS Reports
𝑃 𝑀𝑎𝑙𝑒 ∩ 𝐼𝑚𝑝𝑟𝑜𝑝𝑒𝑟𝑇𝑃𝑆
𝑃 𝐼𝑚𝑝𝑟𝑜𝑝𝑒𝑟𝑇𝑃𝑆 𝑀𝑎𝑙𝑒) =
𝑃(𝑀𝑎𝑙𝑒)
Dependent and Independent events
• Dependent events
– 2 events where the outcome of one is influenced
by the outcome of the other.
• Independent events
– 2 events who’s outcomes are completely
unchanged based on the outcomes of each other.
Dependent and Independent events
• If 𝑃 𝐴 𝐵 = 𝑃 𝐴
• Or if 𝑃 𝐵 𝐴 = 𝑃(𝐵)
• Then the events are independent!
Tossing two Coins
• If A is getting heads on the first toss, and
• If B is getting heads on the second toss,
Are A and B dependent or independent?
Tossing two Coins
• Find 𝑃 𝐵 𝐴
• Find 𝑃 𝐵
Tossing two Coins
• Event A: getting Heads on first toss
• Event B: getting Heads on second toss
• Find 𝑃 𝐵 𝐴
1
𝑃 𝐵∩𝐴
1
4
𝑃 𝐵𝐴 =
=
=
1
𝑃 𝐴
2
2
• Find 𝑃 𝐵
1
𝑃 𝐵 =
2
A useful result!
• Take the mathematical equation of conditional
probability:
𝑃 𝐴∩𝐵
𝑃 𝐵𝐴 =
𝑃 𝐴
Multiply both sides by 𝑃(𝐴)
𝑃 𝐴 𝑃 𝐵 𝐴 =𝑃 𝐴∩𝐵
We can calculate 𝑃(𝐴 ∩ 𝐵) with just 𝑃 𝐴 and 𝑃(𝐵|𝐴).
This is the MULTIPLICATION RULE for intersections.
Even more useful!
• If we think about independent events for a
minute, we can recall that for two events A
and B that are independent:
𝑃 𝐵𝐴 =𝑃 𝐵
But that means…
In our Multiplication Rule for Intersections
𝑃 𝐴 𝑃 𝐵 𝐴 =𝑃 𝐴∩𝐵
If we are looking at ONLY independent events
where 𝑃 𝐵 𝐴 = 𝑃 𝐵
Then, the Multiplication Rule results in
𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 ∗ 𝑃(𝐵)
This works for the probability of more
than 2 events!
• If A, B, C, D, … are all INDEPENDENT events
(coin flips, single die rolls, etc.), then
𝑃 𝐴 ∩ 𝐵 ∩ 𝐶 ∩ 𝐷 ∩. . = 𝑃 𝐴 𝑃 𝐵 𝑃 𝐶 𝑃 𝐷 . .
Ok, take a second…. Breath…
• We just covered a lot with very little
explanation. So, we’re going to do some
examples.
Dependent vs. Independent
Determine whether the events are independent or
dependent
1) A: a coin comes up heads
B: 1d6 comes up even
2) A: 1d6 is thrown and comes up even
B: a second d6 is thrown and the sum of the
two is greater than 4
3) A: An ace is drawn from a standard deck of
cards
B: A second ace is drawn from the same deck
Uh-oh…. Wait a sec.
A: An ace is drawn from a standard deck of
cards
B: A second ace is drawn from the same deck
We can come up with two different answers. It
depends on if we put the first card BACK or NOT!
Replacement
• We talked about it with combinations &
permutations, and now we need to
incorporate it into independent & dependent
events.
Replacement
Namely, if we have replacement, does it make it
dependent or independent?
Take our deck of cards example.
A: An ace is drawn from a standard deck of
cards
B: A second ace is drawn from the same deck
Replacement
If we put the first card back into the deck (and
re-shuffle) then we have an independent event.
The first card has no impact on what the second
card could be.
Replacement
Alternatively, if we leave the first card out, then
the probabilities for the second card have
changed.
One more thing
• Find
𝑃 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 ℎ𝑒𝑎𝑑𝑠 4 𝑡𝑖𝑚𝑒𝑠 𝑜𝑛 𝑎 𝑐𝑜𝑖𝑛 𝑓𝑙𝑖𝑝
• Find 𝑃 𝐻𝐻𝐻𝐻|𝐻𝐻𝐻