Transcript 6.6 - Combined Events - Binomial Probabilities

Binomial Probabilities
IBHL, Y2 - Santowski
(A) Coin Tossing Example
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Take 2 coins and toss each
Make a list to predict the possible outcomes
Determine the ratio in which the possible
outcomes (# of heads) occur
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Now repeat for 3 coins
And repeat again for 4 coins
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So now, predict for 25 coins …… ???
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(A) Coin Tossing Example - Results
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For 2 coins:
2 heads (HH), 1 head (either HT or TH), no heads (TT)
So ratio is 1:2:1
For 3 coins:
3 heads (HHH), 2 heads (either HHT, HTH, THH), 1 head (either TTH, THT,
HTT), 0 head (TTT)
So the ratio is 1:3:3:1
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For 4 coins:
4 heads (HHHH), 3 heads (either HHHT, HHTH, HTHH, THHH), 2 heads
(HHTT, TTHH, HTHT, THTH, THHT, HTTH), 1 head (HTTT, THTT, TTHT,
TTTH) or 0 head (TTTT)
So the ratio is 1:4:6:4:1
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So then, what if we had 25 coins??????
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(A) Coin Tossing Example - Patterns
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To look for a pattern, we return to combinatorials:
Our ratio was 1:4:6:4:1 for 4H, 3H, 2H, 1H, 0H
To relate combinations  our 1 result of 4 heads from 4 coins could be
interpreted as C(4,4) since we are looking for how many different
combinations of 4H we can make from 4 coins  obviously 1 such
combination
Likewise, from the 4 coins, we can look for the number of combinations of
3H  C(4,3) which equals 4
From the 4 coins, we next consider the number of combinations of 2H 
C(4,2) = 6
From the 4 coins, we now consider the number of combinations of 1H 
C(4,1) = 4
And finally, from the 4 coins, the number of combinations of 0 H  C(4,0)
(A) Coin Tossing Example - Generalizations
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So if we have k coins 
Our ratios and meanings would be as follows:
C(k,k)  from k coins, we want k H  k!/k! = 1
C(k,k – 1)  from k coins, we want k – 1 H  k!/[1!x(k – 1)!] = k
C(k,k – 2)  we want k – 2 H  k!/[2!x(k – 2)!] = ½ k(k – 1)
C(k,k – 3)  we want k – 3 H  k!/[3!(k-3)!] = 1/6 [k(k – 1)(k – 2)]
etc….
So for 25 coins and say we wish to get 19 heads  C(25,19) =
25!/(6!19!) = 177,100 ways
(A) Coin Tossing Example – Pascal’s Triangle
& Binomial Expansions
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Notice that our number of possible
ways of getting r Heads from n
coins have ratios identical to the
rows in Pascal’s triangle:
 1
 121
 1331
 14641
 1 5 10 10 5 1
 1 6 15 20 15 6 1
 1 7 21 35 35 21 7 1
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Then let’s also relate Pascal’s
triangle and the observed
ratios to binomial expansions
of (a + b)n
 (a + b)1
 (a + b)2
 (a + b)3
 (a + b)4
 (a + b)5
 (a + b)6
 (a + b)7
(A) Coin Tossing Example – Pascal’s Triangle
& Binomial Expansions - Conclusion
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We notice that the coefficients in our
binomial expansions (or in Pascal’s
Triangle) are identical to the ratios of our
possible outcomes in our coin scenario
(B) Pascal’s Triangle & Binomial Expansions Probabilities
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Now let’s change the scenario slightly 
we will now consider probabilities rather
than simply counting the number of ways
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So let a represent the probability of getting
H with our coin (i.e. p(H) = a = ½ )
Then b will represent the probability of
getting T with our coin (i.e. p(T) = b = ½ )
(B) Pascal’s Triangle & Binomial Expansions Probabilities
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So now a meaning to our binomial expression (a + b)2 would be 
tossing a coin twice (or tossing 2 coins simultaneously)
Then our expansion a2 + 2ab + b2 would relate to the probabilities as
follows:
a2 would be the probability of getting 2 H, which would equal (½)2 = ¼
ab would mean the probability of getting one H and one T, which can
happen in 2 different ways, hence 2ab  which would equal 2( ½ )( ½ )
=½
And b2 would be the probability of getting 2 T, which would equal (½)2 =
¼
(B) Pascal’s Triangle & Binomial Expansions Probabilities
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Now what about 4 coins?
Our starting point would be (a + b)4
The expansion is a4 + 4a3b + 6a2b2 + 4ab3 + b4  The expansion has the following
interpretation:
a4  all 4 coins are heads and the probability is (½)4 = 1/16
a3b  3 of the 4 coins are H (which happens in 4 different combinations) and the
probability is 4(½)3(½) = 4/16
a2b2  2 of the 4 coins are H (which happens in 6 different combinations) and the
probability is 6(½)2(½)2 = 6/16
ab3 1 of the 4 coins are H (which happens in 4 different combinations) and the
probability is 4(½)(½)3 = 4/16
b4  0 of the 4 coins are H (which happens in 1 different combinations) and the
probability is 1(½)(½)4 = 1/16
And how many different outcomes are there  1+4+6+4+1 = 16
(B) Pascal’s Triangle & Binomial Expansions Probabilities
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Now what if our coin was “weighted”  let p(H) = ¾ and p(T) = ¼
Our starting point is again (a + b)4
And the expansion is again a4 + 4a3b + 6a2b2 + 4ab3 + b4  The expansion
has the following interpretation:
a4  all 4 coins are heads and the probability is (¾ )4 = 81/256
a3b  3 of the 4 coins are H (which happens in 4 different combinations)
and the probability is 4(¾)3(¼) = 108/256
a2b2  2 of the 4 coins are H (which happens in 6 different combinations)
and the probability is 6(¾)2(¼)2 = 54/256
ab3 1 of the 4 coins are H (which happens in 4 different combinations)
and the probability is 4(¾)(¼)3 = 12/256
b4  0 of the 4 coins are H (which happens in 1 different combinations) and
the probability is 1(¾)(¼)4 = 1/256
(C) Binomial Probabilities - Generalizations
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Let p equal the probability of “success” or simply our given event happening
Then let q equal the probability of “failure” or “not successful” or simply our
event not happening
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As well, we will run the “experiment” n times and we will look for “success”
or the occurrence of our event in r of these n trials  therefore, the “failure,
non-success” or simply the non-occurrence of our event will occur n – r
times
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Then the probability that the event occurs r times (or is successful r times)
AND that our event does NOT occur (failure) n – r times is:
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P(E=r) = C(n,r) x pr x qn-r
Where P(E=r) is read as “the probability that the event occurs r times”
which other texts will write as P(X = r)
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(C) Binomial Probabilities - Generalizations
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We will make the following 4 assumptions in the working
of this formula:
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(i) we have a fixed number of trials
(ii) the probability for the occurrence of our event (or
success) is the same each time
(iii) each trial is independent of all the other events
(iv) each event only has two possible outcomes 
occurrence or non-occurrence (success of failure)
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(D) Examples
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Ex 1. An archer has a 90% chance of hitting a target with
each arrow. She shoots 5 arrows. Determine the
probability that she hits the target only twice
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Soln #1  Start with (p + q)5 and expand as p5 + 5p4q +
10p3q2 + 10p2q3 + 5pq4 + q5 and simply interpret hits
twice means 2 occurrence (or two successes), so we
need to look for p2 in our expansion  hence the term of
interest is 10p2q3 and our substitution into the formula is
10(0.9)2(0.1)3 = 0.0081
(D) Examples
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Ex 1. An archer has a 90% chance of hitting a target with each arrow.
She shoots 5 arrows. Determine the probability that she hits the
target only twice
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Soln #2  Start with (0.9 + 0.1)5 and expand as (0.9)5 + 5(0.9)4(0.1)
+ 10 (0.9)3(0.1)2 + 10 (0.9)2 (0.1)3 + 5 (0.9) (0.1)4 + (0.1)5 and simply
interpret hits twice means 2 occurrence (or two successes), so we
need to look for (0.9)2 in our expansion  hence the term of interest
is 10(0.9)2(0.1)3 which is 10(0.9)2(0.1)3 = 0.0081
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Soln #3  use the formula P(E=2) = C(5,2) x (0.9)2 x (0.1)3 = 0.0081
(D) Examples
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Ex 1. An archer has a 90% chance of hitting a target with each arrow. She
shoots 5 arrows. Determine the probability that she hits the target only
twice.
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Soln #4 – Use the FCP and combinatorials:
If we have 5 total events, then we set up 5 spaces (one for each arrow shot)
 E1 E2 E3 E4 E5  one possible combination for the 5 arrows could be
H H’ H H’ H’  so how many possible combinations of these events are
there??  C(5,2)
Then the FCP says simply take the product of the 5 probabilities 
(0.9)(0.1)(0.9)(0.1)(0.1) x 10 = 0.0081
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Alternatively, we can use permutations with identical objects  we have 5
“events” , (the arrows), which can be arranged in P(5,5) permutations (120
permutations)  but we have 3 identical objects (3 misses) and 2 other
identical objects (2 hits)  so P(5,5) ÷ (3!2!) = 10
(D) Examples
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Ex 1. An archer has a 90% chance of hitting a target with each arrow. She
shoots 5 arrows. Determine the probability that she hits the target only
twice.
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Soln #5  use the GDC  we can use the distribution key to answer this
question:
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Hit 2nd VARS (to get to the distributions menu)
Scroll down the list of options for the distributions to find 0:binompdf(
Which stands for the binomial probability distribution function (more on that
later)
Hit ENTER to copy the command to your home screen
Syntax is binompdf(5,0.9,2) which represents 5 total arrows being shot,
where success happens at a probability of 0.9 (90% success rate) and we
want to have 2 successes)
Answer on GDC is 0.0081
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(D) Examples
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Ex 1. An archer has a 90% chance of hitting a target with each arrow. She
shoots 5 arrows. Determine the probability that she hits the target at most 3
times.
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We change the question slightly and have to solve for P(E = 0,1,2,3)
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You can use any of the previously demonstrated 5 methods
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NOTE on GDC solution  from the DISTRIBUTIONS menu, you will select
A:binomcdf(  which stands for a cumulative distribution function  as
you are cumulating the probabilities from 0,1,2, and 3 hits
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Answer is 0.08146
(E) Further Examples
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Ex 2: A die is tossed 7 times. Find the probability that:
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(i) exactly 2 tosses give a 6
(ii) at least 2 tosses result in a 6
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ANS (i) 0.234 (ii) 0.330
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Ex 3: A doctor estimates that his treatment of a particular disease is
successful 75% of the time. Find the probability that he will
successfully treat exactly 5 of 6 patients who seek his help.
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ANS = 0.356
(E) Further Examples
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Ex 4: Five percent of a large shipment of fruit is inedible (let’s say
that the shipment was 10,000 bananas). Find the probability that in
a random selection of 10 bananas from the shipment, exactly 2 are
inedible.
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ANS = 0.0746
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Ex 5: How many rolls of a die are required to ensure that the
probability of obtaining at least one “double 6” is greater than 95%
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ANS=107