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

Worked example 1

\(P(A)=0.5\), \(P(B)=0.4\), \(P(A\cup B)=0.7\). Are \(A\) and \(B\) independent?

\(P(A\cap B)=0.5+0.4-0.7=0.2\) and \(P(A)P(B)=0.2\), so \(A\) and \(B\) are independent.

Worked example 2

A bag has 5 red and 3 blue counters; two are taken without replacement. Find \(P(\text{both red})\), \(P(\text{different})\) and \(P(\text{2nd red}\mid\text{1st blue})\).

\(\dfrac58\times\dfrac47=\dfrac5{14}\). Different: \(\dfrac58\times\dfrac37+\dfrac38\times\dfrac57=\dfrac{15}{28}\). Given the first is blue, 5 of the remaining 7 are red: \(\tfrac57\).

Worked example 3

Machine A makes 60% of items, 2% defective; machine B makes 40%, 5% defective. Find \(P(\text{defective})\) and \(P(A\mid\text{defective})\).

\(P(D)=0.6(0.02)+0.4(0.05)=0.032\). Reversing the conditioning, \(P(A\mid D)=\dfrac{0.012}{0.032}=0.375\).

Worked example 4

A test has prevalence 1%, detection rate 95% and false positive rate 4%. Find \(P(\text{disease}\mid\text{positive})\).

\(P(\text{positive})=0.01(0.95)+0.99(0.04)=0.0491\), so \(P(\text{disease}\mid\text{positive})=\dfrac{0.0095}{0.0491}\approx0.194\). The value is low because the condition is rare: a classic exam context.

Worked example 5

\(X\sim N(50,4^2)\). Find \(P(X>56)\), \(P(46\lt X<54)\) and \(a\) with \(P(X\lt a)=0.9\).

\(P(Z>1.5)=0.0668\). \(P(-1\lt Z<1)=0.6827\). Inverse Normal gives \(z=1.2816\), so \(a=50+4(1.2816)=55.13\).

Worked example 6

\(P(X<30)=0.2\) and \(P(X>50)=0.1\). Find \(\mu\) and \(\sigma\).

\(z=-0.8416\) for 30 and \(z=1.2816\) for 50 (since \(P(X<50)=0.9\)). Subtracting, \(\dfrac{20}{\sigma}=2.1232\), so \(\sigma=9.42\) and \(\mu=30+0.8416(9.42)=37.9\).

Worked example 7

Egg masses \(X\sim N(60,5^2)\) g. Find the proportion between 50 g and 68 g, and the probability that all of five eggs are in this range.

\(P(-2\lt Z<1.6)=0.9452-0.0228=0.9224\); \(0.9224^5=0.668\).

Worked example 8

\(X\sim B(100,0.4)\). Estimate \(P(X\ge45)\).

\(Y\sim N(40,24)\). \(P(X\ge45)\approx P(Y>44.5)=P\left(Z>\dfrac{4.5}{\sqrt{24}}\right)=0.179\); the exact binomial value is 0.1789.

Worked example 9: A two-way table with a given condition

In a college, 150 students study maths. 90 are girls. Of the girls, 54 also study physics; of the boys, 36 study physics. Find the probability that a student chosen at random studies physics given that they are a boy.

There are \(150-90=60\) boys. Of these, 36 study physics. The word "given" tells us the denominator is the number of boys, not 150. So \(P(\text{physics}\mid\text{boy})=\dfrac{36}{60}=0.6\). For comparison, \(P(\text{physics}\mid\text{girl})=\dfrac{54}{90}=0.6\) as well, so in this college studying physics does not depend on sex: the events are independent.

Worked example 10: Reading a probability from the Normal curve

The time spent on homework by students is modelled by \(H\sim N(95,20^2)\) minutes. Find the probability that a student spends more than 2 hours.

Two hours is 120 minutes. Standardise: \(z=\dfrac{120-95}{20}=1.25\). Sketch the curve and shade the right tail. \(P(H>120)=1-\Phi(1.25)=1-0.8944=0.1056\). About one student in ten spends over two hours.

Worked example 11: Both parameters unknown with a shortcut

\(X\sim N(\mu,\sigma^2)\) and the quartiles of \(X\) are 40 and 60. Find \(\mu\) and \(\sigma\).

The Normal curve is symmetrical, so the mean lies halfway between the quartiles: \(\mu=50\). The upper quartile has \(z=0.6745\), so \(\dfrac{60-50}{\sigma}=0.6745\) and \(\sigma=14.83\). Always look for symmetry before setting up simultaneous equations.

Worked example 12: At least one in a sample

The mass \(M\) of a parcel is modelled by \(N(2.5,0.6^2)\) kg. A courier will not carry parcels over 3.5 kg. Find the probability that, of 4 parcels, at least one is over the limit.

\(P(M>3.5)=P(Z>1.667)=0.0478\). "At least one" is best found as \(1-P(\text{none})\). \(P(\text{none})=(1-0.0478)^4=0.9522^4=0.822\). So \(P(\text{at least one})=1-0.822=0.178\).

Worked example 13: A tree diagram for a sequence of events

Whether a student passes a driving test at the first attempt is 0.5. A student who fails passes at the second attempt with probability 0.7. Find the probability that a student passes within two attempts, and that a student who passes needed two attempts.

Pass on first attempt: 0.5. Fail then pass: \(0.5\times0.7=0.35\). So \(P(\text{pass within two})=0.5+0.35=0.85\). Given that the student passed, the probability that two attempts were needed is \(\dfrac{0.35}{0.85}=0.412\).

Worked example 14: Judging whether a Normal model is sensible

A researcher records the number of text messages sent in a day by 200 students. The mean is 12 and the standard deviation is 15. Comment on a Normal model.

If the data were Normal, about 95% would lie within two standard deviations, so values between \(12-30=-18\) and \(12+30=42\). A count cannot be negative. The mean is smaller than the standard deviation, which signals strong positive skew. So the Normal model is not appropriate: \(P(X<0)=P(Z<-0.8)=0.21\) would give a 21% chance of an impossible negative number.

Worked example 15: Using symmetry and the Normal curve in context

The lengths of screws made by a machine are \(L\sim N(40,0.5^2)\) mm. Screws shorter than 39 mm or longer than 41 mm are rejected. Find the proportion rejected, and the proportion rejected when the machine is reset so that the standard deviation is 0.3 mm.

The limits are 1 mm either side of the mean. With \(\sigma=0.5\) they are \(\pm2\) standard deviations, so \(P(\text{accepted})=P(-2\lt Z<2)=0.9545\) and \(P(\text{rejected})=0.0455\). With \(\sigma=0.3\) the limits are \(\pm3.33\) standard deviations, so \(P(\text{rejected})=2P(Z>3.333)=2\times0.000429=0.00086\). A smaller standard deviation makes the production far more consistent, even though the mean has not changed.