CIE AS/A level Probability & Statistics 1 Paper 5 Prediction - 2025
CIE AS/A level Probability & Statistics 1 Paper 5 Prediction- 2025
To excel in A level Math Exam, consistent practice with CIE AS & A Level Revision resources is key. CIE AS/A level Probability & Statistics 1 Paper 5 Prediction will guide you for exam pattern.
IITian Academy offers a vast collection of questions that can aid your understanding of specific topics and solidify your concepts. By practicing regularly and focusing on these key areas, you’ll be well-prepared for the A level Math exam
Question 1
In a certain country, the probability of more than 10 cm of rain on any particular day is 0.18, independently of the weather on any other day. Find the probability that in any randomly chosen 7-day period, more than 2 days have more than 10 cm of rain.
Answer/Explanation
Ans:
[P(>2) = 1 – P(0,1,2) =]
1 – (7C0 0.180 0.827 + 7C1 0.181 0.826 + 7C2 0.182 0.825)
= 1 – (0.249285 + 0.383048 + 0.252251)
= 1 – 0.88458
0.115
Question 2
For 3 randomly chosen 7-day periods, find the probability that exactly two of these periods have at least one day with more than 10 cm of rain.
Answer/Explanation
Ans:
[P(at least 1 day of rain) = 1 – P(0) = 1- ( 0.82)7 = ] 0.7507
[P(exactly 2 periods) =] 0.75072 × (1 – 0.7507) × 3
0.421
Question 3
Box A contains 7 red balls and 1 blue ball. Box B contains 9 red balls and 5 blue balls. A ball is chosen at random from box A and placed in box B. A ball is then chosen at random from box B. The tree diagram below shows the possibilities for the colours of the balls chosen.
(a) Complete the tree diagram to show the probabilities.
(b) Find the probability that the two balls chosen are not the same colour.
(c) Find the probability that the ball chosen form box A is blue given that given that the ball chosen form box B is blue
Answer/Explanation
Ans:
(a)
(b) \(\frac{7}{8} \times \frac{5}{15}+\frac{1}{8} \times \frac{9}{15}\)
\(=\frac{44}{120}[\frac{11}{30}or0.367]\)
(c) P(A blue|B blue) \(=\frac{A blue \bigcap B blue)}{P(B blue)}\)
\(=\frac{\frac{1}{8}\times \frac{6}{15}}{\frac{7}{8}\times \frac{5}{15}+\frac{1}{8}\times \frac{6}{15}}=\frac{\frac{1}{20}}{\frac{41}{120}}\)
\(=\frac{6}{41}\) or 0.146
Question 4
A fair eight-sided die has faces marked 1, 2, 3, 4, 5, 6, 7, 8. The score when the die is thrown is the number on the face the die lands on. The die is thrown twice.
- Event R is ‘one of the scores is exactly 3 greater than the other score’.
- Event S is ‘the product of the scores is more than 19’.
(i) Find the probability of R. [2]
(ii) Find the probability of S. [2]
(iii) Determine whether events R and S are independent. Justify your answer. [3]
Answer/Explanation
Ans:
Question 5
The lengths of female snakes of a particular species are normally distributed with mean 54 cm and standard deviation 6.1 cm.
(a) Find the probability that a randomly chosen female snake of this species has length between 50 cm and 60 cm. The lengths of male snakes of this species also have a normal distribution. A scientist measures the lengths of a random sample of 200 male snakes of this species. He finds that 32 have lengths less than 45 cm and 17 have lengths more than 56 cm.
(b) Find estimates for the mean and standard deviation of the lengths of male snakes of this species.
Answer/Explanation
Ans
6 (a) \(P\left ( \frac{50-54}{6.1}< z< \frac{60-54}{6.1} \right )=P(-0.6557< Z< 0.9836)\)
Both values correct
Φ (0.9836) – Φ (–0.6557) = Φ (0.9836) + Φ (0.6557) – 1
= 0.8375 + 0.7441 – 1
(Correct area)
0.582
6 (b) \(\frac{45-\mu }{\sigma }=-0.994\)
\(\frac{56-\mu }{\sigma }=-1.372\)
One appropriate standardisation equation with µ, σ ,z-value (not probability) and 45 or 56.
11 = 2.366σ
(M1 for correct algebraic elimination of µ or σ from their two simultaneous equations to form an equation in one variable)
σ = 4.65, μ = 49.6
Question 6
In Greenton, 70% of the adults own a car. A random sample of 8 adults from Greenton is chosen.
(a) Find the probability that the number of adults in this sample who own a car is less than 6.
A random sample of 120 adults from Greenton is now chosen.
(b) Use an approximation to find the probability that more than 75 of them own a car.
Answer/Explanation
Ans:
(a) 1 – P(6,7,8)
\(=1-(^8C_6 0.7^6 0.3^2 + ^8C_7 0.3^1 +0.7^8)\)
= 1 – 0.55177
= 0.448
Alternative method for question 5(a)
P(0,1,2,3,4,5)
\(=0.3^8 + ^8C_10.7^1 0.3^7 + ^8C_2 0.7^2 0.3^6 +^8C_30.7^3 0.3^5 + ^8C_4 0.7^4 0.3^4 + ^8C_5 0.7^50.3^3\)
=0.448
(b) Mean = \(120 \times 0.7 = 84\)
Var =\(120 \times 0.7 \times 0.3 = 25.2\)
P( more than 75 ) = P\((z>\frac{75.5-84}{\sqrt{25.2}}\)
\(P(z>-1.693)\)
= 0.955
Question 7
(a) Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be arranged so that all 3 Es are together.
(b) Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be arranged so that the Ps are not next to each other.
(c) Find the probability that a randomly chosen arrangement of the 10 letters of the word SHOPKEEPER has an E at the beginning and an E at the end. Four letters are selected from the 10 letters of the word SHOPKEEPER.
(d) Find the number of different selections if the four letters include exactly one P.
Answer/Explanation
Ans
7 (a) \(\frac{8!}{2!}\)
20160
7 (b) Total number of ways: \(\frac{10!}{2!3!}(=302400)(A)\)
With Ps together: \(\frac{9!}{3!}(=60480)(B)\)
With Ps not together: 302 400 – 60 480
241 920
Alternative method for question 7(b)
\(\frac{8!}{3!}\)
\(\times \frac{9\times 8}{2}\)
241 920
7 (c) \(Probability=\frac{Number\ of\ ways\ Es\ at\ beginning\ and\ end}{Total\ number\ of\ ways}\)
\(Probability=\frac{\frac{8!}{2!}}{\frac{10!}{2!\times 3!}}=\frac{20160}{302400}\)
\(\frac{1}{15}, 0·0667 \)
Alternative method for question 7(c)
\(Probability =\frac{3}{10}\times \frac{2}{9}\)
\(\frac{1}{15}, 0·0667 \)
Alternative method for question 7(c)
\(Probability=\frac{1}{10}\times \frac{1}{9}\times 3!\)
\(\frac{1}{15},0.0667 \)
7(d) Scenarios:
P E E E 5C0 = 1
P E E _ 5C1 = 5
P E _ _ 5C2 = 10
P _ _ _ 5C3 = 10
Total = 26