1. Binomial Probabilities IBHL, Y2 - Santowski

2. (A) Coin Tossing Example 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 Now repeat for 3 coins And repeat again for 4 coins So now, predict for 25 coins …… ???

3. (A) Coin Tossing Example - Results 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 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 So then, what if we had 25 coins??????

4. (A) Coin Tossing Example - Patterns 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)

5. (A) Coin Tossing Example - Generalizations 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

6. (A) Coin Tossing Example – Pascal’s Triangle & Binomial Expansions 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 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 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

7. (A) Coin Tossing Example – Pascal’s Triangle & Binomial Expansions - Conclusion 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

8. (B) Pascal’s Triangle & Binomial Expansions - Probabilities Now let’s change the scenario slightly  we will now consider probabilities rather than simply counting the number of ways …. 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 )

9. (B) Pascal’s Triangle & Binomial Expansions - Probabilities 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 a 2 + 2ab + b 2 would relate to the probabilities as follows: a 2 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 b 2 would be the probability of getting 2 T, which would equal (½) 2 = ¼

10. (B) Pascal’s Triangle & Binomial Expansions - Probabilities Now what about 4 coins? Our starting point would be (a + b) 4 The expansion is a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4  The expansion has the following interpretation: a 4  all 4 coins are heads and the probability is (½) 4 = 1/16 a 3 b  3 of the 4 coins are H (which happens in 4 different combinations) and the probability is 4(½) 3 (½) = 4/16 a 2 b 2  2 of the 4 coins are H (which happens in 6 different combinations) and the probability is 6(½) 2 (½) 2 = 6/16 ab 3  1 of the 4 coins are H (which happens in 4 different combinations) and the probability is 4(½)(½) 3 = 4/16 b 4  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

11. (B) Pascal’s Triangle & Binomial Expansions - Probabilities 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 a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4  The expansion has the following interpretation: a 4  all 4 coins are heads and the probability is (¾ ) 4 = 81/256 a 3 b  3 of the 4 coins are H (which happens in 4 different combinations) and the probability is 4(¾) 3 (¼) = 108/256 a 2 b 2  2 of the 4 coins are H (which happens in 6 different combinations) and the probability is 6(¾) 2 (¼) 2 = 54/256 ab 3  1 of the 4 coins are H (which happens in 4 different combinations) and the probability is 4(¾)(¼) 3 = 12/256 b 4  0 of the 4 coins are H (which happens in 1 different combinations) and the probability is 1(¾)(¼) 4 = 1/256

12. (C) Binomial Probabilities - Generalizations 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 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 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: P(E=r) = C(n,r) x p r x q n-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)

13. (C) Binomial Probabilities - Generalizations We will make the following 4 assumptions in the working of this formula: (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)

14. (D) Examples 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 Soln #1  Start with (p + q) 5 and expand as p 5 + 5p 4 q + 10p 3 q 2 + 10p 2 q 3 + 5pq 4 + q 5 and simply interpret  hits twice means 2 occurrence (or two successes), so we need to look for p 2 in our expansion  hence the term of interest is 10p 2 q 3 and our substitution into the formula is 10(0.9) 2 (0.1) 3 = 0.0081

15. (D) Examples 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 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 Soln #3  use the formula P(E=2) = C(5,2) x (0.9) 2 x (0.1) 3 = 0.0081

16. (D) Examples 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. 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 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

17. (D) Examples 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. Soln #5  use the GDC  we can use the distribution key to answer this question: Hit 2 nd 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

18. (D) Examples 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. We change the question slightly and have to solve for P(E = 0,1,2,3) You can use any of the previously demonstrated 5 methods 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 Answer is 0.08146

19. (E) Further Examples Ex 2: A die is tossed 7 times. Find the probability that: (i) exactly 2 tosses give a 6 (ii) at least 2 tosses result in a 6 ANS (i) 0.234 (ii) 0.330 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. ANS = 0.356

20. (E) Further Examples 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. ANS = 0.0746 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% ANS=107