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Binomial Probabilities

Binomial Probabilities. IBHL, Y2 - Santowski. (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

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Binomial Probabilities

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

  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 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 b2would 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 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

  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 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

  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 rof 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 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)

  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 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

  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 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

  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

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