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AS & A Level Mathematics (9709)

Discrete random variables Banker

29 questions · about 1% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 51 · Q1 (a, b)6 marksE(X) and Var(X) · Probability distribution table
  • 2026 May/June · Paper 52 · Q2 (a)3 marksProbability distribution table
  • 2026 May/June · Paper 53 · Q2 (a, b)6 marksE(X) and Var(X) · Probability distribution table
  • 2026 May/June · Paper 55 · Q5 (a, b)6 marksE(X) and Var(X) · Probability distribution table

2025

  • 2025 Oct/Nov · Paper 51 · Q1 (a, b)4 marksE(X) and Var(X) · Probability distribution table
  • 2025 Oct/Nov · Paper 52 · Q2 (a, b)7 marksE(X) and Var(X) · Probability distribution table
  • 2025 Oct/Nov · Paper 53 · Q4 (a, b)6 marksE(X) and Var(X) · Probability distribution table
  • 2025 Oct/Nov · Paper 55 · Q2 (a)4 marksE(X) and Var(X)
  • 2025 May/June · Paper 51 · Q6 (b)4 marksProbability distribution table
  • 2025 May/June · Paper 55 · Q1 (a, b)6 marksE(X) and Var(X) · Probability distribution table

2024

  • 2024 Oct/Nov · Paper 51 · Q2 (a, b)6 marksE(X) and Var(X) · Probability distribution table
  • 2024 Oct/Nov · Paper 52 · Q3 (a, b)6 marksE(X) and Var(X) · Probability distribution table
  • 2024 Oct/Nov · Paper 53 · Q2 (a, b)4 marksE(X) and Var(X) · Probability distribution table
  • 2024 May/June · Paper 51 · Q6 (a)4 marksOfficial paper ↗Probability distribution table
  • 2024 May/June · Paper 52 · Q5 (a, b, c)7 marksE(X) and Var(X) · Probability distribution table
  • 2024 May/June · Paper 53 · Q1 (a, b)6 marksE(X) and Var(X) · Probability distribution table

2023

  • 2023 Oct/Nov · Paper 51 · Q5 (a)1 markProbability distribution table
  • 2023 Oct/Nov · Paper 52 · Q1 (a, b)5 marksE(X) and Var(X) · Probability distribution table
  • 2023 Oct/Nov · Paper 53 · Q1 (a, b)6 marksE(X) and Var(X) · Probability distribution table
  • 2023 May/June · Paper 51 · Q6 (a, b, c)6 marksE(X) and Var(X) · Probability distribution table
  • 2023 May/June · Paper 52 · Q1 (a, b)6 marksE(X) and Var(X) · Probability distribution table
  • 2023 May/June · Paper 52 · Q4 (c)3 marksProbability distribution table
  • 2023 May/June · Paper 53 · Q3 (a, b, c)7 marksE(X) and Var(X) · Probability distribution table

2022

  • 2022 Oct/Nov · Paper 51 · Q14 marksE(X) and Var(X)
  • 2022 Oct/Nov · Paper 52 · Q5 (c)3 marksProbability distribution table
  • 2022 Oct/Nov · Paper 53 · Q4 (a, b)5 marksProbability distribution table
  • 2022 May/June · Paper 51 · Q4 (a, b)5 marksE(X) and Var(X) · Probability distribution table
  • 2022 May/June · Paper 52 · Q2 (a, b)6 marksE(X) and Var(X) · Probability distribution table
  • 2022 May/June · Paper 53 · Q3 (a, b)6 marksE(X) and Var(X) · Probability distribution table

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Probability distribution table · Core · Paper 5 style · 3 marks
  1. (a) The random variable X takes the values 1, 2, 3 and 4 only, and P(X = x) = kx, where k is a constant. Draw up the probability distribution table for X in terms of k and find the value of k. [2]
  2. (b) Find P(X ≥ 3). [1]
Show answer
  1. (a) k = 1/10; P(X = x) = 1/10, 2/10, 3/10, 4/10 for x = 1, 2, 3, 4
    M1 k + 2k + 3k + 4k = 1; A1 k = 1/10.
  2. (b) 7/10
    B1 3/10 + 4/10.
Q2Probability distribution table · Core · Paper 5 style · 5 marks
  1. (a) A fair spinner has six equal sectors labelled 1, 1, 2, 3, 3, 3. The spinner is spun twice and X is the sum of the two scores. Draw up the probability distribution table for X. [4]
  2. (b) Find P(X is even). [1]
Show answer
  1. (a) x: 2, 3, 4, 5, 6; P: 1/9, 1/9, 13/36, 1/6, 1/4
    B1 values 2 to 6; M1 P(1) = 1/3, P(2) = 1/6, P(3) = 1/2 used with products and ×2 for orders; A1 three probabilities correct; A1 all correct (total 1).
  2. (b) 13/18
    B1 1/9 + 13/36 + 1/4 = 26/36.
Q3Probability distribution table · Core · Paper 5 style · 3 marks
  1. (a) The probability distribution of X is given in the table.x: 0, 1, 2, 3P(X = x): 0.1, p, 0.3, 2pFind the value of p. [2]
  2. (b) Find P(X < 2). [1]
Show answer
  1. (a) p = 0.2
    M1 0.1 + p + 0.3 + 2p = 1; A1 p = 0.2.
  2. (b) 0.3
    B1 0.1 + 0.2.
Q4Probability distribution table · Core · Paper 5 style · 4 marks
  1. (a) At a school fair in Multan, a player pays Rs 20 to throw a fair six-sided die once. If the score is 6 the player receives Rs 70; if the score is 4 or 5 the player receives Rs 30; otherwise the player receives nothing. X is the player's net gain in rupees (amount received minus Rs 20). Draw up the probability distribution table for X. [3]
  2. (b) Find P(X > 0). [1]
Show answer
  1. (a) x: 50, 10, −20; P: 1/6, 1/3, 1/2
    B1 values 50, 10, −20; B1 1/6 and 1/3; B1 1/2.
  2. (b) 1/2
    B1 1/6 + 1/3.
Q5Probability distribution table · Core · Paper 5 style · 4 marks
  1. (a) In a charity raffle in Islamabad, 200 tickets are sold. There is 1 prize of Rs 5000, there are 4 prizes of Rs 1000 and 15 prizes of Rs 200; the other tickets win nothing. Each ticket can win at most one prize. X is the prize money, in rupees, won by one randomly chosen ticket. Draw up the probability distribution table for X. [3]
  2. (b) Find the probability that a randomly chosen ticket wins at least Rs 1000. [1]
Show answer
  1. (a) x: 0, 200, 1000, 5000; P: 9/10, 3/40, 1/50, 1/200
    B1 values; B1 three probabilities correct; B1 all correct (180/200 for 0).
  2. (b) 1/40
    B1 5/200.
Q6Probability distribution table · Core · Paper 5 style · 4 marks
  1. (a) The table shows the probability distribution of X.x: −2, −1, 0, 1, 2, 3P(X = x): 0.05, 0.15, 0.2, 0.25, 0.2, qFind the value of q. [1]
  2. (b) Find P(X ≤ 0). [1]
  3. (c) Find P(|X| ≤ 1). [2]
Show answer
  1. (a) q = 0.15
    B1.
  2. (b) 0.4
    B1 0.05 + 0.15 + 0.2.
  3. (c) 0.6
    M1 values −1, 0, 1; A1 0.15 + 0.2 + 0.25.
Q7Probability distribution table · Standard · Paper 5 style · 5 marks
  1. (a) A bag contains 5 red balls and 3 green balls. Three balls are taken at random from the bag without replacement. The random variable X is the number of green balls taken. Show that P(X = 1) = 15/28. [2]
  2. (b) Draw up the probability distribution table for X. [3]
Show answer
  1. (a) 3 × (5/8 × 4/7 × 3/6) = 15/28 (or ³C₁ × ⁵C₂ / ⁸C₃ = 30/56)
    M1 correct product or combinations method; A1 15/28.
  2. (b) x: 0, 1, 2, 3; P: 5/28, 15/28, 15/56, 1/56
    B1 P(X = 0) = 10/56; B1 P(X = 2) = 15/56; B1 P(X = 3) = 1/56.
Q8Probability distribution table · Standard · Paper 5 style · 5 marks
  1. (a) Bilal practises archery. The probability that he hits the target with any shot is 0.6, independently of other shots. He shoots until he hits the target or until he has taken 4 shots, whichever comes first. X is the number of shots taken. Draw up the probability distribution table for X. [4]
  2. (b) Find the probability that Bilal takes more than 2 shots. [1]
Show answer
  1. (a) x: 1, 2, 3, 4; P: 0.6, 0.24, 0.096, 0.064
    B1 P(1) = 0.6; B1 P(2) = 0.4 × 0.6; B1 P(3) = 0.4² × 0.6; B1 P(4) = 0.4³ (he stops after 4 shots whatever happens).
  2. (b) 0.16
    B1 0.096 + 0.064 (or 0.4²).
Q9Probability distribution table · Standard · Paper 5 style · 4 marks
  1. (a) Two fair four-sided dice, each numbered 1, 2, 3, 4, are thrown. X is the larger of the two scores (or the common score if they are equal). Show that P(X = 3) = 5/16. [2]
  2. (b) Draw up the probability distribution table for X. [2]
Show answer
  1. (a) Outcomes (3,1), (3,2), (1,3), (2,3), (3,3): 5 of 16
    M1 count outcomes; A1.
  2. (b) x: 1, 2, 3, 4; P: 1/16, 3/16, 5/16, 7/16
    B1 at least two other probabilities correct; B1 all correct.
Q10Probability distribution table · Standard · Paper 5 style · 5 marks
  1. (a) The random variable X takes the values 1, 2, 3 and 4. P(X = 1) = P(X = 4) = a and P(X = 2) = P(X = 3) = 3a. Find the value of a. [2]
  2. (b) Find P(X is even). [1]
  3. (c) Find P(X = 2 | X ≥ 2). [2]
Show answer
  1. (a) a = 1/8
    M1 8a = 1; A1.
  2. (b) 1/2
    B1 3/8 + 1/8.
  3. (c) 3/7
    M1 (3/8) ÷ (7/8); A1 3/7.
Q11Probability distribution table · Standard · Paper 5 style · 6 marks
  1. (a) Two fair six-sided dice are thrown. The random variable X is the positive difference between the two scores (X = 0 if the scores are equal). Draw up the probability distribution table for X. [4]
  2. (b) Find P(X ≤ 2 | X ≠ 0). [2]
Show answer
  1. (a) x: 0, 1, 2, 3, 4, 5; P: 6/36, 10/36, 8/36, 6/36, 4/36, 2/36
    M1 sample space of 36 outcomes; A1 P(0) and P(1); A1 P(2) and P(3); A1 P(4) and P(5).
  2. (b) 3/5
    M1 (10 + 8)/30; A1 3/5.
Q12Probability distribution table · Standard · Paper 5 style · 5 marks
  1. (a) A box contains 3 red discs and 2 blue discs. Discs are taken at random, one at a time and without replacement, until a blue disc is obtained. X is the number of discs taken. Show that P(X = 3) = 1/5. [2]
  2. (b) Draw up the probability distribution table for X. [3]
Show answer
  1. (a) (3/5)(2/4)(2/3) = 1/5
    M1 product of three correct fractions; A1.
  2. (b) x: 1, 2, 3, 4; P: 2/5, 3/10, 1/5, 1/10
    B1 P(1) = 2/5; B1 P(2) = 3/10; B1 P(4) = 1/10.
Q13Probability distribution table · Challenge · Paper 5 style · 7 marks
  1. (a) The random variable X takes the values −1, 0, 1 and 2 only, with P(X = x) = k(x² + 1). Find the value of k and draw up the probability distribution table for X. [3]
  2. (b) Find P(X² ≤ 1). [1]
  3. (c) Two independent values of X are taken. Find the probability that their sum is 2. [3]
Show answer
  1. (a) k = 1/10; P: 2/10, 1/10, 2/10, 5/10 for x = −1, 0, 1, 2
    M1 k(2 + 1 + 2 + 5) = 1; A1 k = 1/10; A1 table.
  2. (b) 1/2
    B1 (2 + 1 + 2)/10.
  3. (c) 7/50
    M1 identify pairs (0, 2), (2, 0) and (1, 1) (no pair uses −1, since 3 is not a value); M1 2 × (1/10)(5/10) + (2/10)²; A1 10/100 + 4/100 = 7/50.
Q14Probability distribution table · Challenge · Paper 5 style · 6 marks
  1. (a) A committee of 4 people is chosen at random from 6 women and 4 men. X is the number of men on the committee. Draw up the probability distribution table for X. [5]
  2. (b) Find the probability that the committee contains more men than women. [1]
Show answer
  1. (a) x: 0, 1, 2, 3, 4; P: 1/14, 8/21, 3/7, 4/35, 1/210
    M1 denominator ¹⁰C₄ = 210; M1 P(X = x) = ⁴Cₓ × ⁶C₄₋ₓ / 210; A1 15/210 and 80/210; A1 90/210 and 24/210; A1 1/210.
  2. (b) 5/42
    B1 (24 + 1)/210 = 25/210.
Q15Probability distribution table · Challenge · Paper 5 style · 6 marks
  1. (a) Five cards are numbered 1, 2, 3, 4 and 5. Three cards are chosen at random without replacement. The random variable X is the median of the three numbers chosen. Show that P(X = 3) = 2/5. [2]
  2. (b) Draw up the probability distribution table for X. [2]
  3. (c) Find the probability that the largest number chosen is 5, given that the median is 3. [2]
Show answer
  1. (a) Median 3 needs one card from {1, 2} and one from {4, 5}: 2 × 2 = 4 of ⁵C₃ = 10 selections
    M1 count; A1 4/10.
  2. (b) x: 2, 3, 4; P: 3/10, 2/5, 3/10
    B1 P(X = 2) = 1 × 3/10; B1 P(X = 4) = 3/10.
  3. (c) 1/2
    M1 selections {1,3,5}, {2,3,5} out of 4; A1.
Q16Probability distribution table · Challenge · Paper 5 style · 7 marks
  1. (a) Ayesha has three coins. Two are fair and the third is biased so that the probability of a head is 0.7. She tosses all three coins once. X is the number of heads obtained. Draw up the probability distribution table for X. [5]
  2. (b) Given that exactly two heads were obtained, find the probability that the biased coin showed a head. [2]
Show answer
  1. (a) x: 0, 1, 2, 3; P: 0.075, 0.325, 0.425, 0.175
    M1 consider the biased coin separately; A1 P(0) = 0.25 × 0.3; A1 P(1) = 0.5 × 0.3 + 0.25 × 0.7; A1 P(2) = 0.25 × 0.3 + 0.5 × 0.7; A1 P(3) = 0.25 × 0.7.
  2. (b) 14/17 (≈ 0.824)
    M1 0.35 / 0.425; A1.
Q17Probability distribution table · Challenge · Paper 5 style · 5 marks
  1. (a) The random variable X takes the values 1, 2, 3 and 6 only, with P(X = x) = k/x. Find the value of k. [2]
  2. (b) Two independent observations X₁ and X₂ of X are taken. Find P(X₁X₂ = 6). [3]
Show answer
  1. (a) k = 1/2
    M1 k(1 + ½ + ⅓ + ⅙) = 1; A1 2k = 1.
  2. (b) 1/6
    M1 pairs (1, 6), (6, 1), (2, 3), (3, 2); M1 2(½)(1/12) + 2(¼)(1/6); A1 1/12 + 1/12 = 1/6.
Q18Probability distribution table · Challenge · Paper 5 style · 6 marks
  1. (a) Hamza takes two free throws in a basketball match. The probability that he scores with the first throw is 0.7. If he scores with the first throw, the probability that he scores with the second is 0.8; if he misses the first, the probability that he scores with the second is 0.5. Each successful throw scores 1 point. X is the total number of points from the two throws. Draw up the probability distribution table for X. [4]
  2. (b) Given that Hamza scored exactly 1 point, find the probability that he missed the first throw. [2]
Show answer
  1. (a) x: 0, 1, 2; P: 0.15, 0.29, 0.56
    B1 P(0) = 0.3 × 0.5; M1 P(1) = 0.7 × 0.2 + 0.3 × 0.5; A1 0.29; B1 P(2) = 0.56.
  2. (b) 15/29 (≈ 0.517)
    M1 0.15/0.29; A1.
Q19E(X) and Var(X) · Core · Paper 5 style · 4 marks
  1. (a) The random variable X has the probability distribution shown.x: 1, 2, 3, 4P(X = x): 0.1, 0.3, 0.4, 0.2Find E(X). [2]
  2. (b) Find Var(X). [2]
Show answer
  1. (a) 2.7
    M1 Σx p; A1 0.1 + 0.6 + 1.2 + 0.8 = 2.7.
  2. (b) 0.81
    M1 Σx²p − (E(X))² = 8.1 − 7.29; A1 0.81.
Q20E(X) and Var(X) · Core · Paper 5 style · 3 marks
  1. (a) A fair octahedral die has faces numbered 1 to 8. X is the score when the die is thrown once. Find E(X). [1]
  2. (b) Find Var(X). [2]
Show answer
  1. (a) 4.5
    B1 (1 + 2 + … + 8)/8 = 36/8.
  2. (b) 21/4 = 5.25
    M1 Σx²/8 = 204/8 = 25.5; A1 25.5 − 20.25 = 5.25.
Q21E(X) and Var(X) · Core · Paper 5 style · 5 marks
  1. (a) The number of cups of tea, X, that Khalid drinks on a randomly chosen day has the distribution below.x: 0, 1, 2, 3, 4P(X = x): 0.05, 0.25, 0.4, 0.2, 0.1Find E(X). [2]
  2. (b) Find the standard deviation of X, giving your answer correct to 3 significant figures. [3]
Show answer
  1. (a) 2.05
    M1 Σxp; A1.
  2. (b) 1.02
    M1 E(X²) = 5.25; M1 Var = 5.25 − 2.05² = 1.0475; A1 √1.0475 = 1.02.
Q22E(X) and Var(X) · Core · Paper 5 style · 5 marks
  1. (a) The random variable X has the distribution shown.x: −1, 0, 2, 5P(X = x): 1/4, 1/3, 1/4, 1/6Show that E(X) = 13/12. [2]
  2. (b) Find Var(X), giving your answer correct to 3 significant figures. [3]
Show answer
  1. (a) −1/4 + 0 + 1/2 + 5/6 = 13/12
    M1 Σxp; A1.
  2. (b) 611/144 ≈ 4.24
    M1 E(X²) = 1/4 + 1 + 25/6 = 65/12; M1 65/12 − (13/12)²; A1 4.24.
Q23E(X) and Var(X) · Standard · Paper 5 style · 5 marks
  1. (a) The random variable X takes the values 1, 2, 3, 4 with P(X = x) = (x + 1)/14. Show that E(X) = 20/7. [2]
  2. (b) Find Var(X), giving your answer as an exact fraction. [3]
Show answer
  1. (a) (2 + 6 + 12 + 20)/14 = 40/14 = 20/7
    M1 Σx(x + 1)/14; A1.
  2. (b) 55/49
    M1 E(X²) = (2 + 12 + 36 + 80)/14 = 65/7; M1 65/7 − (20/7)²; A1 55/49.
Q24E(X) and Var(X) · Standard · Paper 5 style · 5 marks
  1. (a) The probability distribution of X is given in the table, and E(X) = 1.4.x: 0, 1, 2, 3P(X = x): 0.2, p, q, 0.1Find the values of p and q. [3]
  2. (b) Find Var(X). [2]
Show answer
  1. (a) p = 0.3, q = 0.4
    M1 p + q = 0.7; M1 p + 2q + 0.3 = 1.4; A1 both.
  2. (b) 0.84
    M1 E(X²) = 0.3 + 1.6 + 0.9 = 2.8; A1 2.8 − 1.96 = 0.84.
Q25E(X) and Var(X) · Standard · Paper 5 style · 5 marks
  1. (a) The random variable X takes the values 1, 2, 3, 4 with P(X = x) = kx(5 − x). Find the value of k. [2]
  2. (b) Explain, without calculation, why E(X) = 2.5. [1]
  3. (c) Find Var(X). [2]
Show answer
  1. (a) k = 1/20
    M1 k(4 + 6 + 6 + 4) = 1; A1.
  2. (b) The distribution is symmetrical about 2.5.
    B1.
  3. (c) 1.05
    M1 E(X²) = (4 + 24 + 54 + 64)/20 = 7.3; A1 7.3 − 6.25 = 1.05.
Q26E(X) and Var(X) · Standard · Paper 5 style · 6 marks
  1. (a) Five cards are numbered 1, 2, 2, 3 and 4. Two cards are chosen at random without replacement. X is the sum of the numbers on the two cards. Draw up the probability distribution table for X. [3]
  2. (b) Find E(X) and Var(X). [3]
Show answer
  1. (a) x: 3, 4, 5, 6, 7; P: 0.2, 0.2, 0.3, 0.2, 0.1
    M1 list the 10 equally likely pairs; A1 three probabilities correct; A1 all correct.
  2. (b) E(X) = 4.8; Var(X) = 1.56
    B1 E(X) = 4.8; M1 E(X²) = 24.6; A1 24.6 − 23.04 = 1.56.
Q27E(X) and Var(X) · Challenge · Paper 5 style · 6 marks
  1. (a) In a game, a player throws two fair six-sided dice. If the dice show a double, the player wins Rs 300. If the scores add up to 7, the player wins Rs 150. Otherwise the player wins nothing. W is the amount won, in rupees. Find E(W). [3]
  2. (b) Find Var(W). [2]
  3. (c) The player pays Rs 80 to play. Find the player's expected gain per game and comment. [1]
Show answer
  1. (a) Rs 75
    B1 P(double) = 1/6; B1 P(sum 7) = 1/6 (these events cannot both happen); B1 E(W) = 300/6 + 150/6 = 75.
  2. (b) 13125
    M1 E(W²) = 90000/6 + 22500/6 = 18750; A1 18750 − 75² = 13125.
  3. (c) −Rs 5; on average the player loses Rs 5 per game
    B1.
Q28E(X) and Var(X) · Challenge · Paper 5 style · 7 marks
  1. (a) A shop in Faisalabad has 10 phones in stock, of which 3 are faulty. A customer buys 3 of these phones, chosen at random. X is the number of faulty phones bought. Draw up the probability distribution table for X. [3]
  2. (b) Find E(X) and Var(X). [4]
Show answer
  1. (a) x: 0, 1, 2, 3; P: 35/120, 63/120, 21/120, 1/120 (= 7/24, 21/40, 7/40, 1/120)
    M1 ³Cₓ × ⁷C₃₋ₓ / ¹⁰C₃; A1 two correct; A1 all correct.
  2. (b) E(X) = 0.9; Var(X) = 0.49
    M1 Σxp = 108/120; A1 0.9; M1 E(X²) = 156/120 = 1.3; A1 1.3 − 0.81 = 0.49.
Q29E(X) and Var(X) · Challenge · Paper 5 style · 6 marks
  1. (a) The random variable X has the distribution shown, where a is a constant with a > 2.x: 0, 2, aP(X = x): 0.4, 0.4, 0.2Show that Var(X) = 0.16a² − 0.32a + 0.96. [3]
  2. (b) Given that Var(X) = 4.8, find the value of a. [3]
Show answer
  1. (a) E(X) = 0.8 + 0.2a; E(X²) = 1.6 + 0.2a²; Var(X) = 1.6 + 0.2a² − (0.8 + 0.2a)² = 0.16a² − 0.32a + 0.96
    B1 E(X); B1 E(X²); B1 expand and simplify.
  2. (b) a = 6
    M1 0.16a² − 0.32a − 3.84 = 0 ⇒ a² − 2a − 24 = 0; M1 (a − 6)(a + 4) = 0; A1 a = 6 (a > 2).
Q30E(X) and Var(X) · Challenge · Paper 5 style · 6 marks
  1. (a) A charity in Lahore runs a game in which a player pays Rs 100 to spin a wheel. The probability of landing on gold is 0.05 and the prize is Rs x; the probability of landing on silver is 0.2 and the prize is Rs 200; otherwise there is no prize. The charity wants an expected profit of Rs 25 per game. Find x. [3]
  2. (b) With this value of x, find the standard deviation of the prize money for one game, correct to 3 significant figures. [3]
Show answer
  1. (a) x = 700
    M1 expected prize = 100 − 25 = 75; M1 0.05x + 0.2 × 200 = 75; A1 x = 700.
  2. (b) Rs 164
    M1 E(P²) = 0.05 × 700² + 0.2 × 200² = 32500; M1 Var = 32500 − 75² = 26875; A1 √26875 = 163.9 ≈ 164.

Similar questions

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Probability distribution table Banker27 questions · 9/9 sessions
E(X) and Var(X) Banker22 questions · 9/9 sessions

Subtopics

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Probability distribution table Banker899/9
E(X) and Var(X) Banker599/9

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