(a)Write down an expression, in terms of x, for the probability that a pen taken at random from the box is green.(1)
(b)Given that the probability the pen is green is 0.375, find the value of x.(2)
(Total for Question 1 is 3 marks)
2
A biased spinner can land on red, blue or green only. The probability that the spinner lands on red is 2y. The probability that it lands on blue is 3y. The probability that it lands on green is 0.25.
(a)Form and solve an equation to find the value of y.(3)
(b)Hence write down the probability that the spinner lands on blue.(1)
(Total for Question 2 is 4 marks)
3
A darts player hits the bullseye with probability x on each throw, independently of any other throw. She throws twice. The probability that she hits the bullseye on both throws is 0.36.
(a)Form an equation and solve it to find the value of x.(2)
(b)Find the probability that she hits the bullseye on at least one of the two throws.(2)
(Total for Question 3 is 4 marks)
4
In a class of 45 students, the number who study French is 2x, the number who study Spanish is (x + 7), and the number who study German is (x - 2). Every student studies exactly one of these three languages.
(a)Form and solve an equation to find the value of x.(3)
(b)A student from the class is selected at random. Find the probability that the student studies German.(2)
(Total for Question 4 is 5 marks)
5
In a class of 60 students, the number who study Music only is 2x, the number who study Drama only is (x + 5), the number who study both Music and Drama is x, and the number who study neither is 15.
(a)Form and solve an equation to find the value of x.(3)
(b)A student from the class is selected at random. Find the probability that the student studies Music (Music only or both Music and Drama).(2)
(Total for Question 5 is 5 marks)
6
A biased spinner can land on Win or Lose only. The probability that it lands on Win is (2x + 1)/9. The probability that it lands on Lose is (x + 2)/9.
(a)Form and solve an equation to find the value of x.(3)
(b)Hence find the probability that the spinner lands on Win.(1)
(Total for Question 6 is 4 marks)
7
A biased dice is rolled twice. The probability that it lands on a 6 on each roll is p, where p < 0.5. The probability that it lands on at least one 6 in the two rolls is 0.19.
(a)Show that p2 - 2p + 0.19 = 0.(3)
(b)Hence find the value of p.(3)
(Total for Question 7 is 6 marks)
8
A bag contains n counters. 3 of the counters are pink, the rest are white. Two counters are taken from the bag at random, without replacement. The probability that both counters are pink is 1/15.
(a)Show that n2 - n - 90 = 0.(3)
(b)Hence find the value of n.(2)
(Total for Question 8 is 5 marks)
9
A bag contains n counters. 4 of the counters are red, the rest are yellow. There are more yellow counters than red counters. Two counters are taken from the bag at random, without replacement. The probability that one counter is red and one counter is yellow is 8/15.
(a)Show that n2 - 16n + 60 = 0.(4)
(b)Hence find the number of yellow counters in the bag.(3)
(Total for Question 9 is 7 marks)
10
A biased coin is such that the probability of landing on heads is p, where p > 0.5. The coin is flipped twice. The probability of getting exactly one head is 0.48.
(a)Show that p2 - p + 0.24 = 0.(3)
(b)Hence find the value of p.(3)
(Total for Question 10 is 6 marks)
11
Bag A contains 4 red counters and 6 blue counters. Bag B contains x red counters and 5 blue counters. A counter is taken at random from Bag A and a counter is taken at random from Bag B. The probability that both counters are red is 0.2.
(a)Form and solve an equation to find the value of x.(3)
(b)Hence find the probability that a counter taken at random from Bag B is blue.(2)
(Total for Question 11 is 5 marks)
12
A biased dice lands on a 6 with probability p on each roll, independently of any other roll. The dice is rolled three times. The probability of getting at least one 6 is 0.657.
(a)Show that (1 - p)3 = 0.343.(2)
(b)Hence find the value of p, giving your answer to 3 significant figures.(2)
(Total for Question 12 is 4 marks)
13
A bag contains 3 red counters, 2 blue counters and x green counters. Two counters are taken from the bag at random, without replacement. The probability that both counters are the same colour is 5/18.
(a)Show that 13x2 - 63x + 44 = 0.(4)
(b)Hence find the number of green counters in the bag.(3)
(Total for Question 13 is 7 marks)
14
A biased spinner can land on red or blue only. The probability that it lands on red is p, where p > 0.5. The spinner is spun twice. The probability that both spins give the same colour is 0.55.
(a)Show that 2p2 - 2p + 0.45 = 0.(3)
(b)Hence find the value of p, giving your answer to 3 significant figures.(3)
(Total for Question 14 is 6 marks)
15
A bag contains n red counters and 9 blue counters. A counter is taken at random from the bag, its colour is noted, and it is then replaced. 3 more red counters are then added to the bag. A counter is now taken at random from the new bag. The probability of taking a red counter after the extra counters are added is 0.15 more than the probability of taking a red counter before the extra counters were added.
(a)Show that n2 + 21n - 72 = 0.(5)
(b)Hence find the value of n.(2)
(Total for Question 15 is 7 marks)
Mark scheme · 8.11 Probability Equation Questions
Question 1
(a) B1 x/24 oe
(a) Answer: x/24
(b) M1 x/24 = 0.375 oe, or 0.375 x 24
(b) A1 x = 9 cao
(b) Answer: x = 9
Question 2
(a) M1 2y + 3y + 0.25 = 1 oe
(a) M1 5y = 0.75 (correct rearrangement)
(a) A1 y = 0.15 cao
(a) Answer: y = 0.15
(b) B1 0.45 ft from part (a)
(b) Answer: 0.45
Question 3
(a) M1 x2 = 0.36 oe
(a) A1 x = 0.6 cao (rejecting x = -0.6 since 0 ≤ probability ≤ 1)
(a) Answer: x = 0.6
(b) M1 1 - (1 - 0.6)2 oe, or 2(0.6)(0.4) + 0.62
(b) A1 0.84 cao
(b) Answer: 0.84
Question 4
(a) M1 2x + (x + 7) + (x - 2) = 45 oe
(a) M1 4x + 5 = 45 (correct collection of terms)
(a) A1 x = 10 cao
(a) Answer: x = 10
(b) M1 German count = x - 2 = 8 ft from (a)
(b) A1 8/45 oe cao
(b) Answer: 8/45
Question 5
(a) M1 2x + (x + 5) + x + 15 = 60 oe
(a) M1 4x + 20 = 60 (correct collection of terms)
(a) A1 x = 10 cao
(a) Answer: x = 10
(b) M1 Music total = 2x + x = 3x = 30 ft from (a)
(b) A1 30/60 = 1/2 oe cao
(b) Answer: 1/2
Question 6
(a) M1 (2x + 1)/9 + (x + 2)/9 = 1 oe
(a) M1 3x + 3 = 9 (correct clearing of the denominator)
(a) A1 x = 2 cao
(a) Answer: x = 2
(b) B1 5/9 ft from part (a)
(b) Answer: 5/9
Question 7
(a) M1 1 - (1 - p)2 = 0.19 oe (complement method)
(a) M1 correct expansion (1 - p)2 = 1 - 2p + p2
(a) A1 p2 - 2p + 0.19 = 0 cso, answer printed, all working shown
(a) Answer: p2 - 2p + 0.19 = 0 (shown)
(b) M1 correct use of the quadratic formula (or equivalent method) with a = 1, b = -2, c = 0.19
(b) A1 p = 1.9 or p = 0.1 (both values found)
(b) A1 p = 0.1 cao (selecting the root with p < 0.5)
(b) Answer: p = 0.1
Question 8
(a) M1 (3/n) x (2/(n-1)) = 1/15 oe
(a) M1 correct cross multiplication, 15 x 6 = n(n-1)
(a) A1 n2 - n - 90 = 0 cso, answer printed, all working shown
(a) Answer: n2 - n - 90 = 0 (shown)
(b) M1 (n - 10)(n + 9) = 0 oe, or correct use of the quadratic formula
(b) A1 n = 10 cao (rejecting n = -9 since n > 0)
(b) Answer: n = 10
Question 9
(a) M1 P(red then yellow) + P(yellow then red) = 2 x (4/n) x ((n-4)/(n-1)) oe
(a) M1 8(n-4)/(n(n-1)) = 8/15 oe, set up ready to cross multiply
(a) M1 correct expansion/simplification, e.g. 15(n-4) = n(n-1)
(a) A1 n2 - 16n + 60 = 0 cso, answer printed, all working shown
(a) Answer: n2 - 16n + 60 = 0 (shown)
(b) M1 (n - 10)(n - 6) = 0 oe, or correct use of the quadratic formula
(b) A1 n = 10 or n = 6 (both roots found)
(b) A1 6 cao, with n = 10 selected since yellow (6) must exceed red (4); n = 6 rejected as it gives only 2 yellow counters
(b) Answer: 6 yellow counters
Question 10
(a) M1 2p(1-p) = 0.48 oe
(a) M1 correct expansion and division by 2, p - p2 = 0.24
(a) A1 p2 - p + 0.24 = 0 cso, answer printed, all working shown
(a) Answer: p2 - p + 0.24 = 0 (shown)
(b) M1 (p - 0.6)(p - 0.4) = 0 oe, or correct use of the quadratic formula
(b) A1 p = 0.6 or p = 0.4 (both values found)
(b) A1 p = 0.6 cao (selecting the root with p > 0.5)
(b) Answer: p = 0.6
Question 11
(a) M1 (4/10) x (x/(x+5)) = 0.2 oe
(a) M1 correct rearrangement, e.g. x = 0.5x + 2.5
(a) A1 x = 5 cao
(a) Answer: x = 5
(b) M1 total in Bag B = x + 5 = 10 ft from (a)
(b) A1 5/10 = 1/2 oe cao
(b) Answer: 1/2
Question 12
(a) M1 1 - (1-p)3 = 0.657 oe (complement method)
(a) A1 (1 - p)3 = 0.343 cso, answer printed
(a) Answer: (1 - p)3 = 0.343 (shown)
(b) M1 1 - p = cube root of 0.343
(b) A1 p = 0.300 (3sf) cao
(b) Answer: p = 0.300
Question 13
(a) M1 number of same-colour pairs = 3 + 1 + x(x-1)/2 oe (using combinations for red, blue, green)
(a) M1 total number of pairs = (5+x)(4+x)/2 oe
(a) M1 sets up [8 + x(x-1)] / [(5+x)(4+x)] = 5/18 and cross multiplies correctly
(a) A1 13x2 - 63x + 44 = 0 cso, answer printed, all working shown
(a) Answer: 13x2 - 63x + 44 = 0 (shown)
(b) M1 (13x - 11)(x - 4) = 0 oe, or correct use of the quadratic formula
(b) A1 x = 4 or x = 11/13 (both roots found)
(b) A1 x = 4 cao (rejecting x = 11/13 since the number of counters must be a positive integer)
(b) Answer: 4 green counters
Question 14
(a) M1 P(same colour) = p2 + (1-p)2 oe
(a) M1 correct expansion, p2 + 1 - 2p + p2 = 0.55
(a) A1 2p2 - 2p + 0.45 = 0 cso, answer printed, all working shown
(a) Answer: 2p2 - 2p + 0.45 = 0 (shown)
(b) M1 correct substitution into the quadratic formula with a = 2, b = -2, c = 0.45
(b) A1 p = 0.658 or p = 0.342 (awrt, both values found)
(b) A1 p = 0.658 cao (selecting the root with p > 0.5)
(b) Answer: p = 0.658
Question 15
(a) M1 P(red before) = n/(n+9) oe
(a) M1 P(red after) = (n+3)/(n+12) oe
(a) M1 (n+3)/(n+12) - n/(n+9) = 0.15 oe
(a) M1 correctly combines fractions over a common denominator and simplifies the numerator to 27
(a) A1 n2 + 21n - 72 = 0 cso, answer printed, all working shown
(a) Answer: n2 + 21n - 72 = 0 (shown)
(b) M1 (n + 24)(n - 3) = 0 oe, or correct use of the quadratic formula