Conditional probability and the Normal distribution – Year 13 A level maths worksheet answers to download as PDF

← Back to the worksheet

1. Answer to Question 1

   Method: for a conditional probability use \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\); two events are independent exactly when \(P(A\cap B)=P(A)P(B)\), or equivalently when \(P(A\mid B)=P(A)\).

   1. Divide the probability of the intersection by the probability of the condition. \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.3}{0.5}=0.6\) and \(P(B\mid A)=\dfrac{0.3}{0.6}=0.5\). So \(P(A\mid B)=0.6\) and \(P(B\mid A)=0.5\).
   2. Subtract the intersection so that it is not counted twice. Addition law: \(P(A\cup B)=0.6+0.5-0.3=0.8\). So \(P(A\cup B)=0.8\).
   3. Compare the conditional probability with the unconditional one. \(P(A\mid B)=0.6=P(A)\) (equivalently \(P(A)P(B)=0.6\times0.5=0.3=P(A\cap B)\)), so the events are independent. Knowing that \(B\) has happened does not change the probability of \(A\).

2. Answer to Question 2

   Method: for a conditional probability use \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\); two events are independent exactly when \(P(A\cap B)=P(A)P(B)\), or equivalently when \(P(A\mid B)=P(A)\).

   1. Rearrange the addition law \(P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\). \(P(X\cap Y)=0.4+0.35-0.61=0.14\). So \(P(X\cap Y)=0.14\).
   2. Use the definition of conditional probability with \(Y\) as the condition. \(P(X\mid Y)=\dfrac{0.14}{0.35}=0.4\), which equals \(P(X)=0.4\). The two probabilities are equal.
   3. Equal conditional and unconditional probabilities is the test for independence. Since \(P(X\mid Y)=P(X)\) the events are independent. Check: \(0.4\times0.35=0.14\). So knowing that \(Y\) happened tells us nothing about \(X\).

3. Answer to Question 3

   1. On the second pick there are only 9 counters left, so the second-stage fractions have denominator 9. First pick: R \(\tfrac6{10}\), Y \(\tfrac4{10}\). After a red: R \(\tfrac59\), Y \(\tfrac49\). After a yellow: R \(\tfrac69\), Y \(\tfrac39\). Each pair of branches from the same point adds up to 1.
   2. Multiply along the branches R then R. \(P(RR)=\tfrac6{10}\times\tfrac59=\tfrac{30}{90}=\tfrac13\). So the probability that both are red is \(\tfrac13\).
   3. There are two paths: red then yellow, or yellow then red. Add them. \(P(RY)+P(YR)=\tfrac6{10}\times\tfrac49+\tfrac4{10}\times\tfrac69=\tfrac{24}{90}+\tfrac{24}{90}=\tfrac{48}{90}=\tfrac8{15}\). So the probability of different colours is \(\tfrac8{15}\).
   4. This is read straight from the second branch after Y. After a yellow has gone, 9 counters remain and 6 are red, so \(P=\tfrac69=\tfrac23\). So the answer is \(\tfrac23\).

4. Answer to Question 4

   Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

   1. Standardise: \(z=\dfrac{100-120}{15}=-1.333\). \(P(X<100)=\Phi(-1.333)=0.0912\). So about 9.12% of values are below 100.
   2. Standardise: \(z=\dfrac{150-120}{15}=2\), then use the upper tail. \(P(X>150)=P(Z>2)=1-0.97725=0.0228\). So \(P(X>150)=0.0228\).
   3. Standardise both limits. \(105\) and \(135\) are one standard deviation either side of the mean, so \(P(105\lt X<135)=P(-1\lt Z<1)=0.683\). This is the well-known result that about 68% of values lie within one standard deviation of the mean.
   4. Think about what probability means for a continuous variable. \(X\) is continuous, so the probability of any single exact value is the area of a zero-width strip, which is 0. So for a continuous model we only find probabilities of intervals.

5. Answer to Question 5

   Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

   1. The area to the left of \(a\) is 0.95, so use the inverse Normal with 0.95. Inverse Normal gives \(z=1.6449\), so \(a=120+1.6449\times15=144.7\). So \(a=144.7\) to 1 decimal place.
   2. An area of 0.9 to the right means an area of 0.1 to the left, so \(a\) is below the mean. \(P(X\lt a)=0.1\), so \(z=-1.2816\) and \(a=120-1.2816\times15=100.8\). So \(a=100.8\).
   3. Use the inverse Normal with area 0.75; \(a\) is the upper quartile of \(X\). \(z=0.6745\), so \(a=120+0.6745\times15=130.1\). So \(a=130.1\).

6. Answer to Question 6

   Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

   1. Use the inverse Normal on the standard Normal distribution with area 0.2 to the left. \(z=-0.8416\). The value is negative because 0.2 is less than 0.5.
   2. Standardise 40 with the unknown mean and set it equal to the \(z\)-value from part a). \(\dfrac{40-\mu}{8}=-0.8416\), so \(40-\mu=-6.733\) and \(\mu=46.73\). So the mean is \(\mu=46.73\) to 2 decimal places.
   3. Now that \(\mu\) is known, standardise 50 and use the upper tail. \(P(X>50)=P\left(Z>\dfrac{50-46.73}{8}\right)=P(Z>0.409)=0.341\). So \(P(X>50)\approx0.341\).

7. Answer to Question 7

   1. The first stage is the bus, the second stage is Jaya's arrival; each pair of branches adds up to 1. First branches: on time 0.8, late 0.2. From each, branches: before bell 0.95 / after bell 0.05 (on time) and 0.4 / 0.6 (late). Writing all four outcomes at the ends of the branches helps for parts b) and c).
   2. Add the two paths that end with before the bell. \(P(\text{before bell})=0.8\times0.95+0.2\times0.4=0.76+0.08=0.84\). So the probability that Jaya arrives before the bell is 0.84.
   3. Use \(P(T\mid B)=\dfrac{P(T\cap B)}{P(B)}\) with the path on time then before the bell. \(P(\text{on time}\mid\text{before bell})=\dfrac{0.76}{0.84}=\dfrac{19}{21}=0.905\). So it is very likely that the bus was on time.
   4. Independent days means the probabilities multiply. \(0.8^5=0.328\) (3 s.f.). So the bus is on time all week only about one week in three.

8. Answer to Question 8

   Method: for a conditional probability use \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\); two events are independent exactly when \(P(A\cap B)=P(A)P(B)\), or equivalently when \(P(A\mid B)=P(A)\).

   1. Use the total of the canteen column out of the grand total. \(P(\text{canteen})=\dfrac{70}{120}=\dfrac7{12}=0.583\). So the probability is \(\tfrac7{12}\approx0.583\).
   2. Given canteen, we restrict ourselves to the 70 canteen students. The denominator is the canteen column: \(\dfrac{42}{70}=0.6\). So 60% of canteen users are in Year 13.
   3. Given Year 12, we restrict ourselves to the 60 Year 12 students. The denominator is the Year 12 row: \(\dfrac{28}{60}=\dfrac7{15}=0.467\). So \(P(\text{canteen}\mid\text{Year 12})\approx0.467\).
   4. Compare the conditional probability with the overall probability. \(P(\text{canteen}\mid\text{Year 12})=0.467\ne P(\text{canteen})=0.583\). Equivalently \(P(\text{canteen})P(\text{Y12})=\tfrac7{12}\times\tfrac12=0.292\ne\tfrac{28}{120}=0.233\). The events are not independent. Year 12 students are less likely to use the canteen.

9. Answer to Question 9

   Method: for a conditional probability use \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\); two events are independent exactly when \(P(A\cap B)=P(A)P(B)\), or equivalently when \(P(A\mid B)=P(A)\).

   1. \(P(U)=0.3\times0.9+0.7\times0.2=0.27+0.14=0.41\).
   2. \(P(R\mid U)=\dfrac{0.27}{0.41}=0.659\).
   3. \(P(U^{c})=0.59\) and \(P(R\cap U^{c})=0.3\times0.1=0.03\), so \(P(R\mid U^{c})=\dfrac{0.03}{0.59}=0.0508\).
   4. The probability of rain rises from 0.3 to 0.659 when he carries an umbrella and falls to 0.051 when he does not, so it is a useful but not perfect predictor.

   So taking an umbrella makes rain more than twice as likely, while leaving it at home makes rain very unlikely.

10. Answer to Question 10

    Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

    1. Standardise: \(z=\dfrac{185-172}{7}=1.857\), then use the upper tail. \(P(H>185)=P(Z>1.857)=0.0316\). So about 3.2% of students are taller than 185 cm.
    2. Standardise both limits: \(\dfrac{165-172}{7}=-1\) and \(\dfrac{180-172}{7}=1.143\). \(P(165\lt H<180)=P(-1\lt Z<1.143)=0.8735-0.1587=0.715\). So about 71.5% of students are between 165 cm and 180 cm.
    3. The tallest 5% means an area of 0.95 to the left of the cut-off. \(z=1.6449\), so \(h=172+1.6449\times7=183.5\) cm. So a student needs to be at least 183.5 cm tall.
    4. Multiply the proportion by the number of students. \(0.05\times240=12\) students. So about 12 students are invited.

11. Answer to Question 11

    Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

    1. Standardise: \(z=\dfrac{2.9-3.00}{0.08}=-1.25\). \(P(T<2.9)=P(Z<-1.25)=0.1056\). So about 10.6% of bags are rejected.
    2. A bag is accepted with probability \(1-0.1056\); the 5 bags are independent, so multiply. \((1-0.1056)^5=0.8944^5=0.572\). So the probability that no bag in the box is rejected is 0.572.
    3. Now the area below 2.9 must be 0.01, so find \(z\) by the inverse Normal and rearrange for \(\mu\). \(P(T<2.9)=0.01\) gives \(z=-2.3263\). So \(\dfrac{2.9-\mu}{0.08}=-2.3263\), \(\mu=2.9+0.1861=3.086\) g. So the factory must raise the mean to about 3.086 g.

12. Answer to Question 12

    1. Add the true positives and the false positives. \(P(+)=0.02\times0.9+0.98\times0.07=0.018+0.0686=0.0866\). So about 8.7% of people test positive.
    2. Divide the true positives by all the positives. \(P(C\mid+)=\dfrac{0.018}{0.0866}=0.208\). So only about 21% of people who test positive have the condition.
    3. A negative result for someone without the condition has probability \(0.98\times0.93\). \(P(-)=0.9134\) and \(P(C^{c}\cap-)=0.98\times0.93=0.9114\), so the probability is \(\dfrac{0.9114}{0.9134}=0.998\). So a negative result is correct 99.8% of the time.
    4. Compare part b) with the claim in the newspaper. Only about 21% of positive results are true positives because the condition is rare, so the statement is false. A negative result, however, is very reliable (99.8%). This is why a positive screening result is always followed by a second test.

13. Answer to Question 13

    1. The second probability depends on the first result: after a goal it is 0.8. \(0.8\times0.8=0.64\). So the probability of scoring both is 0.64.
    2. There are two paths with exactly one goal; find each one and add. Score then miss: \(0.8\times0.2=0.16\). Miss then score: \(0.2\times0.6=0.12\). Total \(0.28\). So the probability of exactly one goal is 0.28.
    3. Use conditional probability: the path score then miss divided by the total from part b). \(\dfrac{0.16}{0.28}=\dfrac47=0.571\). So the probability is \(\tfrac47\approx0.571\).
    4. First find the probability of scoring and of missing the second penalty, then use them as the first stage of a new tree. \(P(S_2)=0.64+0.12=0.76\) and \(P(M_2)=0.24\). \(P(S_3)=0.76\times0.8+0.24\times0.6=0.608+0.144=0.752\). So the probability that she scores her third penalty is 0.752.

14. Answer to Question 14

    Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

    1. Standardise 53 and 63: \(z=-1.4\) and \(z=0.6\). The medium grade is what remains. \(P(M<53)=P(Z<-1.4)=0.0808\). \(P(M>63)=P(Z>0.6)=0.2743\). Medium: \(1-0.0808-0.2743=0.6449\). So about 8.1% are small, 64.5% medium and 27.4% large.
    2. The 6 eggs are independent, so raise the probability of medium to the power 6. \(0.6449^6=0.0719\). So the probability that all six are medium is about 0.0719.
    3. The smallest 10% means an area of 0.1 to the left of the cut-off. \(z=-1.2816\), so the cut-off is \(60-1.2816\times5=53.6\) g. So eggs below about 53.6 g are rejected.

15. Answer to Question 15

    1. First branches A 0.5, B 0.3, C 0.2; from each, defective (0.03, 0.04, 0.06) and good.
    2. \(P(D)=0.5\times0.03+0.3\times0.04+0.2\times0.06=0.015+0.012+0.012=0.039\).
    3. \(P(A\mid D)=\dfrac{0.015}{0.039}=\dfrac5{13}=0.385\) and \(P(C\mid D)=\dfrac{0.012}{0.039}=\dfrac4{13}=0.308\).
    4. Machine A produces 38.5% of the defective cases, more than C (30.8%), because it makes many more cases. A high defect rate does not mean the largest number of defects.

16. Answer to Question 16

    Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

    1. Turn each percentage into an area to the left and use the inverse Normal. \(P(T<35)=0.1\) gives \(z=-1.2816\) so \(35-\mu=-1.2816\sigma\). \(P(T>50)=0.15\) gives \(P(T<50)=0.85\) and \(z=1.0364\), so \(50-\mu=1.0364\sigma\). These are two simultaneous equations in \(\mu\) and \(\sigma\).
    2. Subtract the equations to eliminate \(\mu\). Subtract: \(15=(1.0364+1.2816)\sigma=2.318\sigma\), so \(\sigma=6.47\). Then \(\mu=35+1.2816\times6.47=43.29\). So \(\mu\approx43.29\) minutes and \(\sigma\approx6.47\) minutes.
    3. Standardise 30 and 45 with the values from part b). \(P(30\lt T<45)=P(-2.05\lt Z<0.26)=0.5984-0.0202=0.584\). So about 58% of journeys take between 30 and 45 minutes.
    4. The time allowed must have 99% of journeys below it. \(z=2.3263\), so \(t=43.29+2.3263\times6.47=58.3\) minutes. So the manager should allow about 58 minutes.

17. Answer to Question 17

    1. Start with the intersection, then subtract it from each total. Both: 12. Badminton only \(32-12=20\). Squash only \(28-12=16\). Neither \(60-(20+12+16)=12\). The four regions add up to 60.
    2. The condition tells us which total to divide by. \(P(B\mid S)=\dfrac{12}{28}=\dfrac37=0.429\). \(P(S\mid B)=\dfrac{12}{32}=\dfrac38=0.375\). So the two conditional probabilities are different.
    3. Read the numbers from the Venn diagram. \(P(\text{neither})=\dfrac{12}{60}=0.2\). Exactly one sport: \(\dfrac{20+16}{60}=\dfrac{36}{60}=0.6\). So 20% play neither sport and 60% play exactly one.
    4. Compare \(P(B\cap S)\) with \(P(B)P(S)\). \(P(B)P(S)=\dfrac{32}{60}\times\dfrac{28}{60}=0.249\) but \(P(B\cap S)=\dfrac{12}{60}=0.2\). Since these differ, the events are not independent. Members who play squash are less likely to play badminton.

18. Answer to Question 18

    1. \(X\sim B(200,0.9)\). \(np=180\) and \(np(1-p)=200\times0.9\times0.1=18\), so \(X\approx Y\sim N(180,18)\).
    2. \(P(X>190)=P(X\ge191)\approx P(Y>190.5)=P\left(Z>\dfrac{10.5}{\sqrt{18}}\right)=P(Z>2.475)=0.0067\).
    3. The estimate 0.0067 is about twice the exact value 0.0035. The binomial distribution is skewed because \(p=0.9\) is far from 0.5, so the symmetric Normal curve overestimates the upper tail.
    4. The exact probability 0.0035 is below 0.5%, so 200 tickets is acceptable (the Normal estimate would wrongly suggest otherwise).

19. Answer to Question 19

    1. The class boundaries are at \(\mu-\sigma\), \(\mu\), \(\mu+\sigma\). So \(P(R<215)=0.1587\), \(P(215\lt R<255)=0.3413\), \(P(255\lt R<295)=0.3413\), \(P(R>295)=0.1587\). Multiplying by 200 gives 31.7, 68.3, 68.3, 31.7.
    2. The observed values 31, 69, 64, 36 are all within 5 of the expected values, so the model fits well.
    3. \(z=2.3263\) so \(r=255+2.3263\times40=348\) ms.
    4. Waiting times are usually positively skewed with a long right tail, not symmetrical, so a Normal curve would not fit.

20. Answer to Question 20

    Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

    1. Standardise: \(z=\dfrac{420-500}{50}=-1.6\). \(P(L<420)=P(Z<-1.6)=0.0548\). So about 5.5% of batteries are faulty.
    2. A battery is rejected if it is faulty and caught, or good and wrongly rejected; add the two paths. \(P(\text{rejected})=0.0548\times0.95+0.9452\times0.02=0.05206+0.01890=0.07096\). So about 7.1% of batteries are rejected.
    3. Divide the faulty-and-rejected path by the total from part b). \(P(\text{faulty}\mid\text{rejected})=\dfrac{0.05206}{0.07096}=0.734\). So about 73% of rejected batteries are really faulty.
    4. This is the complement of the rejection probability for a faulty battery. \(1-0.95=0.05\). So 5% of faulty batteries get through quality control.

21. Answer to Question 21

    1. Use the addition law with \(P(A\cap B)=P(A)P(B)\). Independent: \(P(A\cup B)=P(A)+P(B)-P(A)P(B)\), so \(0.8=0.5+P(B)-0.5P(B)\), \(0.3=0.5P(B)\), \(P(B)=0.6\). So \(P(B)=0.6\).
    2. First find the intersection from \(P(A\cap B)=P(B\mid A)P(A)\). \(P(A\cap B)=0.2\times0.5=0.1\), so \(0.8=0.5+P(B)-0.1\) and \(P(B)=0.4\). So \(P(B)=0.4\).
    3. Mutually exclusive means \(P(A\cap B)=0\). Mutually exclusive: \(P(A\cup B)=P(A)+P(B)\), so \(P(B)=0.3\). Then \(P(A\cap B)=0\) but \(P(A)P(B)=0.15\ne0\), so they are not independent. Two events with non-zero probabilities can never be both mutually exclusive and independent.

22. Answer to Question 22

    Method: standardise with \(Z=\dfrac{X-\mu}{\sigma}\), sketch the curve to decide which tail you need, then use the Normal distribution functions of your calculator. For an unknown value, use the inverse Normal function and rearrange \(x=\mu+z\sigma\).

    1. Write the two percentages as areas to the left: 0.2 below 38 and 0.9 below 78. \(z_1=-0.8416\) and \(z_2=1.2816\). Then \(38-\mu=-0.8416\sigma\) and \(78-\mu=1.2816\sigma\). Subtracting: \(40=2.1232\sigma\), \(\sigma=18.84\), \(\mu=38+0.8416\times18.84=53.86\). So \(\mu\approx53.86\) and \(\sigma\approx18.84\).
    2. If 70% pass, then 30% score below the pass mark. \(P(M\lt m)=0.3\) gives \(z=-0.5244\), so \(m=53.86-0.5244\times18.84=43.98\approx44\). So the pass mark is 44.
    3. Each student passes with probability 0.7 independently, so use \(B(3,0.7)\): \(\binom32=3\) arrangements. Binomial: \(3\times0.7^2\times0.3=0.441\). So the probability that exactly two pass is 0.441.