Probability and random variables – university Year 1 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Because \(A\) and \(B\) are independent, \(P(A\cap B)=P(A)P(B)=0.4\times0.35=0.14\) (1). By the addition rule, \(P(A\cup B)=0.4+0.35-0.14=0.61\) (1).
   2. If \(A\) and \(B\) are independent, so are \(A\) and \(B^c\); hence \(P(A\mid B^c)=P(A)=0.4\) (1). By De Morgan, \(P(A^c\cap B^c)=1-P(A\cup B)=1-0.61=0.39\) (1), which also equals \(P(A^c)P(B^c)=0.6\times0.65\), confirming that the complements are independent too.

   In (a), a common error is to add \(0.4+0.35=0.75\) without subtracting the overlap. In (b), the direct calculation \(P(A\cap B^c)\div P(B^c)=0.26\div0.65=0.4\) gives the same result.

2. Answer to Question 2 (4 marks)

   1. Machine 2 makes the remaining 30% of the bolts. By the law of total probability, \(P(D)=0.7\times0.04+0.3\times0.10=0.028+0.030=0.058\) (2), so about 5.8% of all bolts are defective.
   2. By Bayes' theorem, \(P(M_2\mid D)=\dfrac{P(M_2)P(D\mid M_2)}{P(D)}=\dfrac{0.030}{0.058}\approx0.517\) (2). Although machine 2 makes fewer bolts, its higher defect rate means it produces just over half of the defective ones.

   In (a), a tree diagram with two stages, machine then quality, shows the two paths to a defective bolt. In (b), the denominator must be the total \(P(D)\).

3. Answer to Question 3 (4 marks)

   1. \(E[X]=1(0.2)+2(0.5)+3(0.3)=0.2+1+0.9=2.1\). Next \(E[X^2]=1(0.2)+4(0.5)+9(0.3)=0.2+2+2.7=4.9\), so \(\operatorname{Var}(X)=E[X^2]-(E[X])^2=4.9-4.41=0.49\) (2).
   2. By linearity, \(E[5-2X]=5-2(2.1)=5-4.2=0.8\) (1). Adding a constant does not change the spread and the factor is squared: \(\operatorname{Var}(5-2X)=(-2)^2\times0.49=1.96\) (1). Note that the variance is multiplied by 4, not by \(-2\): a variance can never be negative, and the standard deviation doubles from 0.7 to 1.4.

   In (a), check that the probabilities add up to \(0.2+0.5+0.3=1\). A common error is to write \(\operatorname{Var}(X)=4.9-2.1\), forgetting to square the mean.

4. Answer to Question 4 (4 marks)

   1. With \(n=20\) and \(p=0.1\), \(P(X=2)=\binom{20}2(0.1)^2(0.9)^{18}=190\times0.01\times0.1501\approx0.285\) (2).
   2. The Poisson approximation uses \(\lambda=np=20\times0.1=2\), so \(P(Y=2)=\dfrac{2^2e^{-2}}{2!}=2e^{-2}\approx0.271\) (1). The answers are close, but the approximation works best when \(n\) is large and \(p\) small; here \(n=20\) is small, so it is only rough (1). The relative error is about 5%, which would shrink for a larger sample with the same mean.

   In (a), \(0.9^{18}\approx0.1501\) must be kept to enough significant figures.

5. Answer to Question 5 (4 marks)

   1. Standardise both limits: \(\dfrac{29.5-30}{0.4}=-1.25\) and \(\dfrac{30.6-30}{0.4}=1.5\) (1). Then \(P(\text{accepted})=\Phi(1.5)-\Phi(-1.25)=0.9332-0.1056=0.8275\) (1).
   2. Assuming parts are independent, \(P(\text{all 10 accepted})=0.8275^{10}\approx0.151\) (1). Only about 15% of boxes are perfect, so the process needs a smaller spread, or a centre moved towards the middle of the tolerance interval (1). The interval is not symmetric about 30, so centring the process at 30.05 mm would also help.

Total : 20 marks