Probability and the binomial distribution – Year 12 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Use the addition rule, subtracting the intersection so that it is not counted twice. \(P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.4-0.2=\) 0.7. So \(P(A\cup B)=0.7\).
   2. \(A'\cap B'\) means neither \(A\) nor \(B\) happens, which is the complement of \(A\cup B\). Neither event happens: \(1-P(A\cup B)=1-0.7=\) 0.3.
   3. Use the definition of conditional probability. \(P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.2}{0.4}=\) 0.5. Knowing that \(B\) has happened does not change the probability of \(A\).
   4. Two events are independent when \(P(A\cap B)=P(A)\times P(B)\). \(P(A)P(B)=0.5\times0.4=0.2=P(A\cap B)\), so they are independent (also \(P(A|B)=P(A)\)).

2. Answer to Question 2

   1. Fill in the Venn diagram from the middle outwards: start with the 9 who play both. Football only \(22-9=13\); tennis only \(18-9=9\); both 9. Total playing \(=13+9+9=31\), so neither \(=40-31=\) 9. So 9 students play neither sport.
   2. Football only is the part of \(F\) outside \(T\). \(\dfrac{13}{40}=\) 0.325. The probability is 0.325.
   3. Given \(F\), the sample space is reduced to the 22 football players, of whom 9 also play tennis. \(P(T|F)=\dfrac{9}{22}=\) 0.409.
   4. Exactly one sport means football only or tennis only. Exactly one sport: \(13+9=22\), probability \(\dfrac{22}{40}=\) 0.55. The probability is 0.55.

3. Answer to Question 3

   1. Without replacement, the probabilities on the second branches change because one counter has gone. First: \(R\) with \(\tfrac58\), \(B\) with \(\tfrac38\). After a red: \(R\) \(\tfrac47\), \(B\) \(\tfrac37\). After a blue: \(R\) \(\tfrac57\), \(B\) \(\tfrac27\). The branches from each point add up to 1.
   2. Multiply along the branches \(R\) then \(R\). \(P(RR)=\dfrac58\times\dfrac47=\dfrac{20}{56}=\) \(\dfrac5{14}\). The probability that both are red is \(\dfrac5{14}\).
   3. Different colours means red then blue, or blue then red; add the two paths. \(P(RB)+P(BR)=\dfrac58\times\dfrac37+\dfrac38\times\dfrac57=\dfrac{15}{56}+\dfrac{15}{56}=\dfrac{30}{56}=\) \(\dfrac{15}{28}\). The probability is \(\dfrac{15}{28}\).
   4. At least one blue is the complement of both red. \(1-P(RR)=1-\dfrac5{14}=\) \(\dfrac9{14}\). The probability is \(\dfrac9{14}\).

4. Answer to Question 4

   1. Mutually exclusive events cannot happen together, so \(P(X\cap Y)=0\) and the probabilities simply add. \(P(X\cup Y)=0.3+0.5=\) 0.8.
   2. For independent events \(P(X\cap Y)=P(X)P(Y)\); then use the addition rule. \(P(X\cap Y)=0.3\times0.5=0.15\), so \(P(X\cup Y)=0.8-0.15=\) 0.65.
   3. Mutually exclusive means \(P(X\cap Y)=0\). Independent means \(P(X\cap Y)=P(X)P(Y)>0\). These contradict each other. So if both probabilities are non-zero, the events cannot be both mutually exclusive and independent.

5. Answer to Question 5

   1. Use \(P(X=r)=\binom nrp^r(1-p)^{n-r}\) with \(n=8\), \(p=0.3\), \(r=2\). \(\binom82(0.3)^2(0.7)^6=28\times0.09\times0.1176=\) 0.2965. So \(P(X=2)=0.2965\) to 4 decimal places.
   2. \(P(X\le1)\) is the sum of \(P(X=0)\) and \(P(X=1)\). \(P(X=0)+P(X=1)=0.7^8+8\times0.3\times0.7^7=0.0576+0.1977=\) 0.2553. So \(P(X\le1)=0.2553\).
   3. Use the complement: \(P(X\ge3)=1-P(X\le2)\), with \(P(X\le2)=P(X\le1)+P(X=2)\). \(1-P(X\le2)=1-(0.2553+0.2965)=\) 0.4482. So \(P(X\ge3)=0.4482\).
   4. For a binomial distribution \(E(X)=np\) and \(\text{Var}(X)=np(1-p)\). \(E(X)=np=\) 2.4; \(\text{Var}(X)=npq=8\times0.3\times0.7=\) 1.68.

6. Answer to Question 6

   1. Yes: \(X\sim B(10,\tfrac16)\) (fixed number of independent trials, two outcomes, constant probability).
   2. Not exactly: the probability changes after each card is removed, so the trials are not independent. (It is a hypergeometric situation.)
   3. No: the number of trials is not fixed; it is geometric.
   4. A fixed number \(n\) of trials; each trial has two outcomes (success/failure); the trials are independent; the probability of success \(p\) is the same for each trial.

7. Answer to Question 7

   1. The three days are independent, so multiply the probabilities of a dry day. \(0.7^3=\) 0.343. The probability of three dry days is 0.343.
   2. The number of rainy days in 5 days is \(B(5,0.3)\). \(X\sim B(5,0.3)\): \(\binom52(0.3)^2(0.7)^3=10\times0.09\times0.343=\) 0.3087. The probability of exactly 2 rainy days is 0.3087.
   3. At least one rainy day is the complement of 7 dry days. \(1-0.7^7=1-0.0824=\) 0.9176. It is very likely (about 92%) to rain at least once in a week.
   4. The expected value of \(B(n,p)\) is \(np\). Dry days \(\sim B(30,0.7)\), mean \(30\times0.7=\) 21. About 21 dry days are expected in a 30-day month.

8. Answer to Question 8

   1. There are 20 independent bulbs, each faulty or not, with the same probability 0.04. \(X\sim B(20,0.04)\), where \(X\) is the number of faulty bulbs in a box.
   2. No faulty bulbs means all 20 are good. \(0.96^{20}=\) 0.442. The probability is about 0.442.
   3. Add \(P(X=0)\), \(P(X=1)\) and \(P(X=2)\) using the binomial formula. \(P(X\le2)=0.96^{20}+20(0.04)(0.96)^{19}+190(0.04)^2(0.96)^{18}=0.4420+0.3683+0.1458=\) 0.956. The probability of at most 2 faulty bulbs is about 0.956.
   4. More than 2 is the complement of at most 2. \(P(X>2)=1-0.9561=\) 0.0439. \(E(X)=20\times0.04=\) 0.8. About 4.4% of boxes lead to a refund, and on average there are 0.8 faulty bulbs per box.

9. Answer to Question 9

   1. The first set of branches shows whether a person has the disease; the second shows the test result. First branch: disease 0.02, no disease 0.98. Second branch: positive 0.95 (if disease), 0.03 (if no disease).
   2. A positive test can come from a person with the disease or from a healthy person; add the two paths. \(P(+)=0.02\times0.95+0.98\times0.03=0.019+0.0294=\) 0.0484. So about 4.84% of people test positive.
   3. Use \(P(D|+)=\dfrac{P(D\cap+)}{P(+)}\). \(P(D|+)=\dfrac{0.019}{0.0484}=\) 0.393. Only about 39% of people who test positive actually have the disease.
   4. The disease is rare, so most positive results come from the large healthy group even though the false-positive rate is small: 0.0294 of the population against 0.019 genuinely ill.

10. Answer to Question 10

    1. There are 12 independent questions, each correct with probability \(\tfrac14\). \(X\sim B(12,0.25)\). This is the number of correct answers.
    2. Use the binomial formula with \(r=3\). \(\binom{12}3(0.25)^3(0.75)^9=220\times0.015625\times0.07508=\) 0.258. The probability of exactly 3 correct answers is about 0.258.
    3. Passing means 6 or more correct; use the complement of \(P(X\le5)\). \(P(X\ge6)=1-P(X\le5)=1-0.9456=\) 0.0544 (about 5.4%). So she has only about a 5% chance of passing by guessing.
    4. The expected value of \(B(n,p)\) is \(np\). \(E(X)=12\times0.25=\) 3. She expects 3 marks, half the pass mark, so passing by guessing is unlikely.

11. Answer to Question 11

    1. The number of late days in 5 days is \(B(5,0.2)\); never late means 0 late days. \(0.8^5=\) 0.328. The probability is about 0.328.
    2. Use the binomial formula with \(r=2\). \(\binom52(0.2)^2(0.8)^3=10\times0.04\times0.512=\) 0.2048. The probability is 0.2048.
    3. At least 3 means exactly 3, 4 or 5; add the three probabilities. \(P(X\ge3)=\binom53(0.2)^3(0.8)^2+\binom54(0.2)^4(0.8)+(0.2)^5=0.0512+0.0064+0.00032=\) 0.05792. The probability is about 0.0579.
    4. Multiply the probability of a letter in one week by the number of weeks. \(38\times0.05792=2.2\), so about 2 letters. About 2 letters would be expected in a school year.

12. Answer to Question 12

    1. The number of baskets is \(B(10,0.7)\); use the binomial formula with \(r=8\). \(\binom{10}8(0.7)^8(0.3)^2=45\times0.0576\times0.09=\) 0.2335. The probability of exactly 8 baskets is about 0.2335.
    2. Use the complement of \(P(X\le7)\), from a calculator or tables. \(P(X\ge8)=1-P(X\le7)=1-0.6172=\) 0.3828. The probability is about 0.383.
    3. Use the cumulative binomial function on a calculator. \(P(X\le6)=\) 0.3504. The probability is about 0.350.
    4. For \(B(n,p)\) the mean is \(np\) and the variance is \(np(1-p)\). Mean \(=10\times0.7=\) 7. Variance \(=10\times0.7\times0.3=2.1\), standard deviation \(=\sqrt{2.1}=\) 1.45.

13. Answer to Question 13

    1. There are \(6\times6=36\) equally likely outcomes; the totals of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). 6 outcomes out of 36: \(\dfrac16\). The probability is \(\dfrac16\).
    2. At least one six is the complement of no sixes, and \(P(\text{no six})=\tfrac56\times\tfrac56\). \(1-\left(\tfrac56\right)^2=\dfrac{11}{36}\). The probability is \(\dfrac{11}{36}\).
    3. Given at least one six, the sample space is reduced to 11 outcomes; count those with a total of at least 10. With at least one six there are 11 outcomes. Total \(\ge10\) among them: (6,4), (4,6), (6,5), (5,6), (6,6): 5. So \(\dfrac5{11}\). The conditional probability is \(\dfrac5{11}\).
    4. Let \(A\) be at least one six and \(B\) a total of at least 10; compare \(P(A\cap B)\) with \(P(A)P(B)\). \(P(\text{total}\ge10)=\dfrac6{36}=\dfrac16\) (the outcomes (4,6), (5,5), (5,6), (6,4), (6,5), (6,6)). \(P(A)P(B)=\dfrac{11}{36}\times\dfrac16=\dfrac{11}{216}\), but \(P(A\cap B)=\dfrac5{36}=\dfrac{30}{216}\). Not equal, so not independent.

14. Answer to Question 14

    1. Without replacement, the probabilities change after each sock is taken; multiply along the path. \(\dfrac6{10}\times\dfrac59\times\dfrac48=\dfrac{120}{720}=\) \(\dfrac16\). The probability that all three are red is \(\dfrac16\).
    2. Exactly two green socks can happen in three orders, and each order has the same probability. Three orders (GGR, GRG, RGG), each with probability \(\dfrac4{10}\times\dfrac39\times\dfrac68=\dfrac{72}{720}=0.1\). Total \(=3\times0.1=\) 0.3. The probability is 0.3.
    3. At least one green is the complement of all three red. \(1-\dfrac16=\) \(\dfrac56\). The probability is \(\dfrac56\).
    4. The probability of green changes as socks are removed (no replacement), so the trials are not independent. So the conditions for a binomial model are not satisfied.

15. Answer to Question 15

    1. Read the totals from the table and divide by 150. \(P(\text{bus})=\dfrac{60}{150}=0.4\). \(P(\text{Y13}\cap\text{car})=\dfrac{25}{150}=\dfrac16\). So \(P(\text{bus})=0.4\) and \(P(\text{Y13 and car})=\tfrac16\).
    2. Given Year 13, only the 80 Year 13 students are considered. \(\dfrac{25}{80}=\) 0.3125. So 31.25% of Year 13 students come by car.
    3. Given walk, only the 50 walkers are considered. \(\dfrac{30}{50}=\) 0.6. So 60% of walkers are in Year 12.
    4. Independence requires \(P(\text{Y13}\cap\text{car})=P(\text{Y13})\times P(\text{car})\). \(P(\text{Y13})\times P(\text{car})=\dfrac{80}{150}\times\dfrac{40}{150}=0.142\), but \(P(\text{Y13}\cap\text{car})=0.167\). They differ, so the events are not independent: Year 13 students are more likely to come by car.

16. Answer to Question 16

    1. At least one advert is the complement of no advert in \(n\) independent opens. \(P(\text{none})=0.9^n\), so \(P(\ge1)=1-0.9^n\). For \(n=20\): \(1-0.1216=0.8784\) ✓. The entry in the table is correct.
    2. Solve the inequality with logarithms; dividing by \(\ln0.9<0\) reverses the sign. \(1-0.9^n>0.99\Rightarrow0.9^n<0.01\Rightarrow n>\dfrac{\ln0.01}{\ln0.9}=43.7\). So \(n=\) 44.
    3. Substitute \(n=43\) and \(n=44\) into \(1-0.9^n\). \(n=43\): \(1-0.9^{43}=0.9892\) (below 0.99). \(n=44\): \(0.9903\) (above) ✓. So 44 is the smallest number of opens.
    4. The number of adverts is \(B(44,0.1)\), with mean \(np\). \(44\times0.1=\) 4.4. On average 4.4 adverts would be shown in 44 opens.

17. Answer to Question 17

    1. An item is defective if it comes from one of the machines and is defective; add the three products from the table. \(P(D)=0.005+0.006+0.008=\) 0.019. So 1.9% of the factory's output is defective.
    2. Use conditional probability: \(P(C|D)=\dfrac{P(C\cap D)}{P(D)}\). \(P(C|D)=\dfrac{0.008}{0.019}=\) 0.421. So about 42% of defective items come from machine \(C\), even though it makes only 20% of the output.
    3. Assuming the two items are independent, multiply the probabilities. (Independent items) \(0.019^2=\) 0.000361. The probability is 0.000361.
    4. Multiply the batch size by the probability of a defect. \(5000\times0.019=\) 95. About 95 defective items are expected.

18. Answer to Question 18

    1. Read \(P(X\le4)\) directly; for a single value subtract consecutive cumulative probabilities. \(P(X\le4)=\) 0.2173. \(P(X=6)=P(X\le6)-P(X\le5)=0.6098-0.4032=\) 0.2066.
    2. \(X\ge8\) is the complement of \(X\le7\). \(P(X\ge8)=1-P(X\le7)=1-0.7869=\) 0.2131. The probability is 0.2131.
    3. Look for the first cumulative probability above 0.5. \(P(X\le5)=0.4032<0.5\) and \(P(X\le6)=0.6098>0.5\), so \(c=\) 6.
    4. For \(B(n,p)\) the mean is \(np\) and the variance is \(np(1-p)\). Mean \(=15\times0.4=6\); standard deviation \(=\sqrt{15\times0.4\times0.6}=\sqrt{3.6}=1.90\). The value \(c=6\) is the median and it equals the mean, as expected for a roughly symmetrical distribution.

19. Answer to Question 19

    1. Use the inclusion-exclusion principle for three sets. Union \(=34+28+20-12-9-8+5=58\). None \(=60-58=\) 2. So 2 members do none of the sports.
    2. Fill in the Venn diagram from the centre outwards: all three, then each pair only, then each sport only. All three: 5. Swim and cycle only: \(12-5=7\); swim and run only: \(9-5=4\); cycle and run only: \(8-5=3\). Swim only: \(34-7-4-5=18\); cycle only: \(28-7-3-5=13\); run only: \(20-4-3-5=8\). Exactly one: \(18+13+8=\) 39. So 39 members do exactly one sport.
    3. Given \(R\), only the 20 runners are considered; 8 of them also cycle. \(P(C|R)=\dfrac{8}{20}=\) 0.4. So 40% of runners also cycle.
    4. Exactly two sports means one of the three \"pair only\" regions. Exactly two: \(7+4+3=14\); probability \(\dfrac{14}{60}=\) 0.233. The probability is \(\dfrac7{30}\approx0.233\).

20. Answer to Question 20

    1. Use \(E(X)=np\) and \(\text{Var}(X)=npq\); dividing the variance by the mean gives \(q\). \(np=6\) and \(npq=4.2\), so \(q=\dfrac{4.2}6=0.7\), \(p=0.3\) and \(n=\dfrac6{0.3}=\) 20. So \(X\sim B(20,0.3)\).
    2. Use the binomial formula with \(n=20\), \(p=0.3\), \(r=6\). \(\binom{20}6(0.3)^6(0.7)^{14}=38760\times0.000729\times0.0067822=\) 0.1916. So \(P(X=6)\approx0.192\).
    3. \(\dfrac{\binom n{r+1}p^{r+1}q^{n-r-1}}{\binom nrp^rq^{n-r}}=\dfrac{n-r}{r+1}\times\dfrac pq\). With \(n=20\), \(p/q=\tfrac37\): ratio \(=\dfrac{3(20-r)}{7(r+1)}\). It exceeds 1 when \(60-3r>7r+7\), i.e. \(r<5.3\), so the probabilities increase up to \(r=6\) and then fall. For \(r=5\): ratio \(=\dfrac{45}{42}>1\); for \(r=6\): \(\dfrac{42}{49}<1\). So \(P(X=6)\) is the largest: the mode is 6.