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

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1. Answer to Question 1 (4 marks)

   1. Rearrange the addition rule: \(P(A\cap B)=P(A)+P(B)-P(A\cup B)\). \(0.6+0.5-0.8=0.3\) (2) So \(P(A\cap B)=0.3\). Since \(P(A\cap B)\gt 0\), the events can happen together, so they are not mutually exclusive.
   2. Use \(P(B|A)=\dfrac{P(A\cap B)}{P(A)}\) and compare with \(P(B)\). \(P(B|A)=\dfrac{0.3}{0.6}=0.5=P(B)\) (1), so independent (1) Equivalently \(P(A)P(B)=0.6\times0.5=0.3=P(A\cap B)\). So \(A\) and \(B\) are independent: knowing that \(A\) happened does not change the probability of \(B\).

   The second mark in (b) requires the comparison with \(P(B)\), not just the word ‘independent’.

2. Answer to Question 2 (4 marks)

   Same colour means both white or both black. Without replacement the second probabilities change, so multiply along each branch and add the two paths.

   \(P(WW)=\dfrac4{10}\times\dfrac39=\dfrac{12}{90}\) (1). \(P(BB)=\dfrac6{10}\times\dfrac59=\dfrac{30}{90}\) (1). Sum \(=\dfrac{42}{90}\) (1) \(=\dfrac7{15}\) (1).

   The probability that the two balls are the same colour is \(\dfrac7{15}\approx0.467\).

   Alternatively, \(1-P(\text{different colours})=1-2\times\dfrac4{10}\times\dfrac69=1-\dfrac{48}{90}=\dfrac{42}{90}\) ✓.

   Using \(\tfrac4{10}\times\tfrac4{10}\) (with replacement) is the most frequent error and scores only the addition mark.

3. Answer to Question 3 (4 marks)

   Use \(P(X=r)=\binom{10}r(0.2)^r(0.8)^{10-r}\). For (b), use the complement: \(P(X\ge2)=1-P(X=0)-P(X=1)\).

   (a) \(\binom{10}3(0.2)^3(0.8)^7=0.2013\) (2). (b) \(1-0.8^{10}-10(0.2)(0.8)^9=1-0.1074-0.2684=0.6242\) (2).

   So \(P(X=3)\approx0.201\) and \(P(X\ge2)\approx0.624\).

   In (a), \(\binom{10}3=120\), so the probability is \(120\times0.008\times0.2097\). In (b), the complement avoids adding nine separate probabilities.

   The binomial PD and CD functions of a calculator give the same values. Writing \(P(X\ge2)=1-P(X\le2)\) is a common mistake that wrongly removes \(P(X=2)\). Award 1 mark for the method and 1 for the value in each part.

4. Answer to Question 4 (4 marks)

   1. There are 50 independent tosses with constant probability 0.5; use \(np\) and \(np(1-p)\). \(X\sim B(50,0.5)\), mean 25, variance 12.5 (2) The standard deviation is \(\sqrt{12.5}\approx3.54\). The distribution is symmetrical about the mean 25.
   2. Use the binomial formula with \(r=25\). \(\binom{50}{25}0.5^{50}=0.1123\) (2) Even the most likely number of heads has only about an 11% chance.

   No normal approximation is needed here: the exact binomial value is expected.

5. Answer to Question 5 (4 marks)

   1. At least once is the complement of never: \(P(\text{no blue})=0.85^{12}\). \(1-0.85^{12}=0.858\) (2) The probability is about 0.858.
   2. The number of blues is \(B(12,0.15)\); use the binomial formula with \(r=2\). \(\binom{12}2(0.15)^2(0.85)^{10}=0.2924\) (1) The probability is about 0.292.
   3. Solve \(1-0.85^n\gt 0.95\) with logarithms; dividing by \(\ln0.85\lt 0\) reverses the inequality. \(0.85^n\lt 0.05\), \(n\gt 18.4\), so 19 (1) At least 19 spins are needed.

   Check (c): \(0.85^{18}\approx0.0536\) is too large, but \(0.85^{19}\approx0.0456\lt0.05\) ✓.

Total : 20 marks