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A-Level Discrete random variables
What the A-Level syllabus expects for Discrete random variables, and how to practise it.
What the syllabus expects
- What discrete random variables, their probability distributions, expectations and variances are
- The binomial distribution B(n, p) as a discrete probability model, together with the conditions that make it appropriate
- Using the mean and variance of the binomial distribution
Scope: quoted without proof - Deriving the cumulative distribution function belonging to a discrete random variable
How it's examined
Questions on this topic most often ask you to find, show, determine, evaluate. About 8% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
For X with P(X = x) = (1/70)(x^2 + x), x = 1 to 5, two independent observations X_1 and X_2 are taken; find the probability their difference is at least 3.
Show the worked answer
P(X=x)=(x²+x)/70: P(1)=2/70, P(2)=6/70, P(3)=12/70, P(4)=20/70, P(5)=30/70. Need P(|X1-X2|>=3), i.e. difference 3 or 4 (ordered pairs). Diff 4: (1,5),(5,1): 2*(2/70)(30/70)=120/4900. Diff 3: (1,4),(4,1): 2*(2/70)(20/70)=80/4900; (2,5),(5,2): 2*(6/70)(30/70)=360/4900. Total = (120+80+360)/4900 = 560/4900 = 4/35.
Example 2 (2 marks)
In game two (4 red, 6 blue, with replacement, 20 recorded draws), find the probability that the 8th draw is the 6th time blue is recorded.
Show the worked answer
P(blue) = 6/10 = 0.6, P(red) = 0.4, draws independent (with replacement). '8th draw is the 6th blue' means exactly 5 blues in the first 7 draws AND the 8th draw is blue. P = C(7,5)(0.6)⁵(0.4)² * 0.6 = C(7,5)(0.6)⁶(0.4)² = 21 * 0.046656 * 0.16 = 0.1568 (4dp).
Example 3 (2 marks)
In game two (4 red, 6 blue, drawn with replacement, 20 draws with colours recorded), find the probability red is recorded more than 4 but at most 8 times.
Show the worked answer
Let R = number of reds recorded, R ~ B(20, 0.4) (p = 4/10, with replacement). Require P(4 < R <= 8) = P(5 <= R <= 8) = P(R <= 8) - P(R <= 4) = 0.59564 - 0.05095 = 0.54468.
More worked questions on this topic
- For the bag of 4 red and 6 blue balls, game two replaces each drawn ball before the next; a pla (2 marks)
- For the wardrobe scoring (sum if formalities match, product if not), give the probability distr (3 marks)
- With C ~ N(8, 0.2^2) and T ~ N(17, 1.4^2) independent, find the probability her total cycling t (3 marks)
- X~B(16, p) is the number, out of 16, of pralines above the recommended sugar level. The single (3 marks)
- Now p = 5. Boxes are packed 12 per carton; a carton is accepted if at most 1 box is seasonal, e (3 marks)
- The game now uses a biased six-sided die marked 2, 3, 4, 5, 6 and 7. For this die P('3')=p, P(' (3 marks)
- A bag holds 6 red, 6 green, 6 yellow and 6 blue discs, identical except for colour. Four discs (3 marks)
- A bag has 6 red, 6 green, 6 yellow and 6 blue discs (24 in all). Four are drawn without replace (3 marks)
- Given she is late with probability 0.215 when leaving at 6:48 am, and she attends five days wee (4 marks)
More A-Level H2 Maths topics
Functions · Graphs and their transformations · Equations and inequalities · Sequences and series · Vectors in two and three dimensions: basic properties · Scalar and vector products · all of A-Level H2 Maths