S4.1Discrete distributions, including the discrete uniform and the binomial
Edexcel A level Maths (9MA0) · Statistics › Statistical distributions
Practise Discrete distributions, including the discrete uniform and the binomial. 10 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
A student is choosing probability models for five random variables.
(a) Tick (✓) two boxes to show which of these random variables could be modelled by a discrete uniform distribution.[2]
The score when a fair ten-sided dice, numbered 1 to 10, is rolled once
The number of heads when three fair coins are tossed
The last digit of the number on a ticket chosen at random from 1000 tickets numbered 000 to 999
The mass of a randomly chosen apple
The number of sixes when a fair six-sided dice is rolled 10 times
(b) Choose one of the random variables that you did not tick and give a reason why a discrete uniform distribution is not a suitable model for it.[1]
Show the answer and mark scheme
(a)Answer: The score when a fair ten-sided dice, numbered 1 to 10, is rolled once; The last digit of the number on a ticket chosen at random from 1000 tickets numbered 000 to 999
B1 for the score on the ten-sided dice
B1 for the last digit of the ticket number
Worked solution: Each score 1 to 10 on a fair dice is equally likely. Each digit 0 to 9 is the last digit of exactly 100 of the 1000 tickets, so the ten last digits are equally likely. The numbers of heads and of sixes are binomial (their values are not equally likely) and mass is continuous.
(b)Answer: e.g. the number of heads: the values 0, 1, 2, 3 are not equally likely (1 and 2 heads are more likely than 0 or 3).
B1 for a correct reason for a correctly identified variable, e.g. the number of heads takes values 0 to 3 that are not equally likely; the mass is continuous, not discrete; the number of sixes is binomial, with 0, 1 or 2 sixes much more likely than 10
Worked solution: For example, the number of heads has \(P(0) = \frac{1}{8}\) but \(P(1) = \frac{3}{8}\), so its values are not equally likely.
Question 2Medium6 marks
Priya models \(D\), the day of the month on which a randomly chosen person was born, by a discrete uniform distribution over the integers 1 to 31.
(a) Using Priya's model, find \(P(D \ge 29)\).[1]
(b) Explain why Priya's model is not suitable.[1]
(c) A better model assumes that each of the 365 days of a non-leap year is equally likely to be a person's birthday. Using this model, find \(P(D \ge 29)\).[2]
(d) Using the model in the previous part, find \(P(D = 31 \mid D \ge 29)\).[2]
Show the answer and mark scheme
(a)Answer: \(\frac{3}{31} \approx 0.0968\)
B1 for \(\frac{3}{31}\) (or awrt 0.0968)
Worked solution: There are 3 values, 29, 30 and 31, each with probability \(\frac{1}{31}\), so \(P(D \ge 29) = \frac{3}{31}\).
(b)Answer: Not every month has a 29th, 30th or 31st day, so these days are less likely than the others.
B1 for the days 29, 30 and 31 do not occur in every month, so they are less likely than days 1 to 28 (the values are not equally likely)
Worked solution: Days 1 to 28 occur in every month, but the 29th and 30th do not occur in February (in most years) and the 31st occurs in only 7 months. So the 31 values are not equally likely.
(c)Answer: \(\frac{29}{365} \approx 0.0795\)
M1 for counting the days: \(11 + 11 + 7 = 29\)
A1 for \(\frac{29}{365}\) (or awrt 0.0795)
Worked solution: The 29th and the 30th each occur in 11 months (every month except February) and the 31st occurs in 7 months. That is \(11 + 11 + 7 = 29\) days of the year, so \(P(D \ge 29) = \frac{29}{365}\).
(d)Answer: \(\frac{7}{29} \approx 0.241\)
M1 for \(\dfrac{\frac{7}{365}}{\frac{29}{365}}\)
A1 for \(\frac{7}{29}\) (or awrt 0.241)
Worked solution: \(P(D = 31 \mid D \ge 29) = \dfrac{P(D = 31)}{P(D \ge 29)} = \dfrac{\frac{7}{365}}{\frac{29}{365}} = \frac{7}{29}\).
Question 3Hard12 marks
Anya spins a fair spinner numbered 1 to 5 and Ben spins a fair spinner numbered 1 to 6. The player whose spinner shows the higher number wins the game. If the two numbers are equal, the game is a draw.
(a) Show that the probability that Ben wins a game is \(\frac{1}{2}\).[2]
(b) Find the probability that Anya wins a game.[2]
(c) Given that a game is not a draw, find the probability that Anya wins.[2]
(d) Anya and Ben play 12 games. Exactly 3 of the games are draws. Find the probability that Anya wins at least 5 of the 12 games, giving your answer to 4 decimal places.[3]
(e) Ben suggests that the game would be fairer if Anya used a fair spinner numbered 2 to 6 instead. Find the probability that Anya wins a game with this spinner, and comment on Ben's suggestion.[3]
Show the answer and mark scheme
(a)Answer: Shown
M1 for counting the outcomes in which Ben's number is higher: \(5 + 4 + 3 + 2 + 1 = 15\) out of 30
A1* for \(\frac{15}{30} = \frac{1}{2}\) (cso)
Worked solution: There are \(5 \times 6 = 30\) equally likely outcomes. If Anya scores \(a\), Ben wins with any of the \(6 - a\) numbers above \(a\), giving \(5 + 4 + 3 + 2 + 1 = 15\) outcomes. So \(P(\text{Ben wins}) = \frac{15}{30} = \frac{1}{2}\).
(b)Answer: \(\frac{1}{3}\)
M1 for \(P(\text{draw}) = \frac{5}{30}\) or counting \(0 + 1 + 2 + 3 + 4 = 10\) outcomes
A1 for \(\frac{1}{3}\) (or awrt 0.333)
Worked solution: The game is a draw in 5 of the 30 outcomes, so \(P(\text{Anya wins}) = 1 - \frac{15}{30} - \frac{5}{30} = \frac{10}{30} = \frac{1}{3}\).
(c)Answer: \(\frac{2}{5}\)
M1 for \(\dfrac{\frac{1}{3}}{\frac{5}{6}}\) or \(\frac{10}{25}\)
A1 for \(\frac{2}{5}\) (or 0.4)
Worked solution: \(P(\text{Anya wins} \mid \text{not a draw}) = \dfrac{\frac{10}{30}}{\frac{25}{30}} = \frac{2}{5}\).
(d)Answer: \(0.2666\)
M1 for \(W \sim B(9, 0.4)\), where \(W\) is the number of the 9 games that are not draws that Anya wins (using the previous part)
M1 for \(P(W \ge 5) = 1 - P(W \le 4)\)
A1 for awrt 0.2666
Worked solution: The 9 games that are not draws are independent, and each is won by Anya with probability \(\frac{2}{5}\) (from the previous part). So the number of them that Anya wins is \(W \sim B(9, 0.4)\), and \(P(W \ge 5) = 1 - P(W \le 4) = 1 - 0.7334 = 0.2666\).
(e)Answer: \(\frac{1}{2}\); Anya now has the advantage (Ben wins with probability \(\frac{1}{3}\)), so the game is still not fair.
M1 for counting the outcomes in which Anya's number is higher: \(1 + 2 + 3 + 4 + 5 = 15\) out of 30
A1 for \(\frac{1}{2}\)
B1 for a comment: Ben now wins with probability \(\frac{1}{3}\), so the game is still unfair (now in Anya's favour)
Worked solution: If Anya scores \(a\) (from 2 to 6), she wins against the \(a - 1\) lower numbers on Ben's spinner: \(1 + 2 + 3 + 4 + 5 = 15\) of the 30 outcomes, so \(P(\text{Anya wins}) = \frac{1}{2}\). Ben wins in \(4 + 3 + 2 + 1 + 0 = 10\) outcomes, a probability of \(\frac{1}{3}\). The change simply swaps the advantage to Anya, so the game is still not fair.